#$&* course PHY 202 3/17 about 1:25 pm 006. `query 5
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Given Solution: Bernoulli's Equation can be written • 1/2 rho v1^2 + rho g y1 + P1 = 1/2 rho v2^2 + rho g y2 + P2 If altitude is constant then y1 = y2, so that rho g y1 is the same as rho g y2. Subtracting this quantity from both sides, these terms disappear, leaving us • 1/2 rho v1^2 + P1 = 1/2 rho v2^2 + P2. The difference in pressure is P2 - P1. If we subtract P1 from both sides, as well as 1/2 rho v2^2, we get • 1/2 rho v1^2 - 1/2 rho v2^2 = P2 - P1. Thus • change in pressure = P2 - P1 = 1/2 rho ( v1^2 - v2^2 ). Caution: (v1^2 - v2^2) is not the same as (v1 - v2)^2. Convince yourself of that by just picking two unequal and nonzero numbers for v1 and v2, and evaluating both sides. ALTERNATIVE FORMULATION Assuming constant rho, Bernoulli's Equation can be written 1/2 rho `d(v^2) + rho g `dy + `dP = 0. If altitude is constant, then `dy = 0 so that 1/2 rho `d(v^2) + `dP = 0 so that `dP = - 1/2 rho `d(v^2). Caution: `d(v^2) means change in v^2, not the square of the change in v. So `d(v^2) = v2^2 - v1^2, not (v2 - v1)^2. STUDENT SOLUTION: The equation for this situation is Bernoulli's Equation, which as you note is a modified KE+PE equation. Considering ideal conditions with no losses (rho*gy)+(0.5*rho*v^2)+(P) = 0 g= acceleration due to gravity y=altitude rho=density of fluid v=velocity P= pressure Constant altitude causes the first term to go to 0 and disappear. (0.5*rho*v^2)+(P) = constant So here is where we are: Since the altitude h is constant, the two quantities .5 rho v^2 and P are the only things that can change. The sum 1/2 `rho v^2 + P must remain constant. Since fluid velocity v changes, it therefore follows that P must change by a quantity equal and opposite to the change in 1/2 `rho v^2. MORE FORMAL SOLUTION: More formally we could write • 1/2 `rho v1^2 + P1 = 1/2 `rho v2^2 + P2 and rearrange to see that the change in pressure, P2 - P1, must be equal to the change 1/2 `rho v2^2 - 1/2 `rho v1^2 in .5 rho v^2: • P2 - P1 = 1/2 `rho v2^2 - 1/2 `rho v1^2 = 1/2 rho (v2^2 - v1^2). ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: query billiard experiment Do you think that on the average there is a significant difference between the total KE in the x direction and that in the y direction? Support your answer. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I do not think there is a significant difference. The variance that I observed was within 10% and within the given standard deviation. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** In almost every case the average of 30 KE readings in the x and in the y direction differs between the two directions by less than 10% of either KE. This difference is not statistically significant, so we conclude that the total KE is statistically the same in bot directions. ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: What do you think are the average velocities of the 'red' and the 'blue' particles and what do you think it is about the 'blue' particle that makes is so? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Since the dark blue particle is 4 times the mass of the green particle, m1*v1^2 = m2*v2^2 (1/4)*apprx.(5)^2 = 1*v2^2 V2^2 = ‘sqrt(25/4) = 2.5 Half the speed but 4 times the mass would equate to the same KE. The same could be done for the red particle with 64 times the mass and get approx. 5*0.125 = 0.625 speed. The KE again would be approx. equal. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Student answer with good analogy: I did not actually measure the velocities. the red were much faster. I would assume that the blue particle has much more mass a high velocity impact from the other particles made very little change in the blue particles velocity. Similar to a bycycle running into a Mack Truck. INSTRUCTOR NOTE: : It turns out that average kinetic energies of red and blue particles are equal, but the greater mass of the blue particle implies that it needs less v to get the same KE (which is .5 mv^2) ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: What do you think is the most likely velocity of the 'red' particle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 5 * 0.125 = 0.625 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If you watch the velocity display you will see that the red particles seem to average somewhere around 4 or 5 ** Your Self-Critique:The red particle moves slower than any other particle in this experiment. I do not feel it is anywhere near 4 or 5 as noted above. Your Self-Critique Rating:3 ********************************************* Question: If the simulation had 100 particles, how long do you think you would have to watch the simulation before a screen with all the particles on the left-hand side of the screen would occur? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Probability for one particle being in one half of the screen would be 1/2. For 100 particles it would be 1 / (2^100) = 7.9*10^-31. I had to look at the solution below to figure out the rest but I now understand that this means in would happen once in 2^100 frames, or a really long time. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT ANSWER: Considering the random motion at various angles of impact.It would likely be a very rare event. INSTRUCTOR COMMENT This question requires a little fundamental probability but isn't too difficult to understand: If particle position is regarded as random the probability of a particle being on one given side of the screen is 1/2. The probability of 2 particles both being on a given side is 1/2 * 1/2. For 3 particles the probability is 1/2 * 1/2 * 1/2 = 1/8. For 100 particlles the probability is 1 / 2^100, meaning that you would expect to see this phenomenon once in 2^100 screens. If you saw 10 screens per second this would take about 4 * 10^21 years, or just about a trillion times the age of the Earth. In practical terms, then, you just wouldn't expect to see it, ever. ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: prin phy and gen phy problem 10.36 15 cm radius duct replentishes air in 9.2 m x 5.0 m x 4.5 m room every 16 minutes; how fast is air flowing in the duct? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Volume of room = 9.2 m * 5.0 m * 4.5 m = 207 m^3 (207 m^3 / 16 min) * ( 1 min / 60 s) = 0.216 m^3 / s This is the amount that needs replenished per second through the duct. Area of duct = ‘pi * 0.15 m ^2 = 0.07 m^2 0.216 m^3/s / (0.07 m^2) = 3.1 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The volume of the room is 9.2 m * 5.0 m * 4.5 m = 210 m^3. This air is replentished every 16 minutes, or at a rate of 210 m^3 / (16 min * 60 sec/min) = 210 m^3 / (960 sec) = .22 m^3 / second. The cross-sectional area of the duct is pi r^2 = pi * (.15 m)^2 = .071 m^2. The speed of the air flow and the velocity of the air flow are related by rate of volume flow = cross-sectional area * speed of flow, so speed of flow = rate of volume flow / cross-sectional area = .22 m^3 / s / (.071 m^2) = 3.1 m/s, approx. Your Self-Critique:OK ` Your Self-Critique Rating:OK ********************************************* Question: prin phy and gen phy problem 10.40 What gauge pressure is necessary to maintain a firehose stream at an altitude of 15 m?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I knew that velocities would have to be equal but I read the solution below to understand that we need to assume the velocity in the hose to be negligible. This gives ‘rho*g*h1 + P1 = ‘rho*g*h2 + P2 P2 - P1 = ‘rho*g*h1 - ‘rho*g*h2 ‘dP = - ‘rho*g*’dh = -(1000 kg/m^3 * -9.8 m/s^2)*15 m = 147,000 N/m^2 This means that the pressure at h=0 is 147,000 N/m^2 greater than atmospheric pressure. So 248,300 Pa. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** We use Bernoulli's equation. Between the water in the hose before it narrows to the nozzle and the 15m altitude there is a vertical change in position of 15 m. Between the water in the hose before it narrows to the nozzle and the 15 m altitude there is a vertical change in position of 15 m. Assuming the water doesn't move all that fast before the nozzle narrows the flow, and noting that the water at the top of the stream has finally stopped moving for an instant before falling back down, we see that we know the two vertical positions and the velocities (both zero, or very nearly so) at the two points. All that is left is to calculate the pressure difference. The pressure of the water after its exit is simply atmospheric pressure, so it is fairly straightforward to calculate the pressure inside the hose using Bernoulli's equation. Assuming negligible velocity inside the hose we have change in rho g h from inside the hose to 15 m height: `d(rho g h) = 1000 kg/m^3 * 9.8 m/s^2 * 15 m = 147,000 N / m^2, approx. Noting that the velocity term .5 `rho v^2 is zero at both points, the change in pressure is `dP = - `d(rho g h) = -147,000 N/m^2. Since the pressure at the 15 m height is atmospheric, the pressure inside the hose must be 147,000 N/m^2 higher than atmospheric. ** Your Self-Critique:See critique in solution. Your Self-Critique Rating:3 ********************************************* Question: Openstax: The heart of a resting adult pumps blood at a rate of 5.00 L/min. Convert this to cm^3 /s . What is this rate in m^3 /s ? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: (5 L / 1 min) * (1000 mL / 1 L) * (1 cm^3 / 1 mL) * ( 1 min / 60 s) = 83.3 cm^3/s (5 L / 1 min) * (1000 mL / 1 L) * (1 cm^3 / 1 mL) * ((1 m)^3 / (100 cm)^3) * ( 1 min / 60 s) = 8.3*10^-5 m^3/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A cube 10 cm on a side is a liter. It takes 10 layers each of 10 rows each of 10 such cubes to fill a one-meter cube. It follows that l liter is a volume of 10 cm * 10 cm * 10 cm = 1000 cm^3, and a cubic meter is a volume equivalent to 10 * 10 * 10 liters = 1000 liters. 5 L / min is therefore the same as 5 L * (1000 cm^3 / L) / min = 5000 cm^3 / min. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: Openstax: Every few years, winds in Boulder, Colorado, attain sustained speeds of 45.0 m/s (about 100 mi/h) when the jet stream descends during early spring. Approximately what is the force due to the Bernoulli effect on a roof having an area of 220 m^2 ? Typical air density in Boulder is 1.14 kg/m^3 , and the corresponding atmospheric pressure is 8.89×104 N/m^2 . (Bernoulli’s principle as stated in the text assumes laminar flow. Using the principle here produces only an approximate result, because there is significant turbulence.) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: There is minimal change in height so the ‘rho*g*h will not change and can be dropped from Bernoulli’s equation to make .5*‘rho*v1^2 + P1 = .5*’rho*v2^2 + P2 where the 1 variables are inside and the 2 variables are outside the roof. The air is not moving inside and the pressure is standard atmospheric pressure given at this location. P2-P1 = .5*’rho*v1^2 - .5*’rho*v2^2 ‘dP = -.5*’rho * ‘dv^2 = -.5 * 1.14 kg/m^3 * (45 m/s)^2 = -1154.25 N/m^2 Since P2 is obviously higher the change is 1154 N/m^2. P = F/A so F = P*A = 1154 N/m^2 * 220 m^2 = 253,880 N of force exerted on the roof by the wind. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Bernoulli's Equation says that rho g y + 1/2 rho v^2 + P = constant. On one side of the roof v = 0 (that would be the inside). On the other side v = 45 m/s. The difference in y from one side of the roof to the other is small, so that the change in rho g y from one side to the other is negligible. Thus 1/2 rho v^2 + P is constant, and we conclude that 1/2 rho v_1^2 + P_1 = 1/2 rho v_2^2 + P_2, where v_1 and P_1 are on one side of the roof (let's say that's the inside) and v_2 and P_2 are on the other. The pressure change from inside to outside is thus P_2 - P_1 = 1/2 rho v_1^2 - 1/2 rho v_2^2 = 1/2 rho ( v_1^2 - v_2^2) = 1/2 * 1.14 kg/m^3 ( (0 m/s)^2 - (45 m/s)^2) = -1200 kg m^2 / (s^2 m^3) = -1200 kg / (m s^2) = -1200 N / m^2, or -1200 Pa, roughly. Exerted on a roof area of 220 m^2 this pressure will thus exert a force, acting from inside toward outside, of F = P * A = 1200 Pa * 220 m^2 = 1200 N/m^2 * 220 m^2 = 260 000 Newtons, very approximately. Your Self-Critique:???Looking at the given solution, are the v2 and v1 reversed and shouldn’t the result be negative as I showed in my solution? Do we then reason like I did that the P outside is greater and assume it to be positive???
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Given Solution: Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: Openstax: A sump pump (used to drain water from the basement of houses built below the water table) is draining a flooded basement at the rate of 0.750 L/s, with an output pressure of 3.00×10^5 N/m^2 . (a) The water enters a hose with a 3.00-cm inside diameter and rises 2.50 m above the pump. What is its pressure at this point? (b) The hose goes over the foundation wall, losing 0.500 m in height, and widens to 4.00 cm in diameter. What is the pressure now? You may neglect frictional losses in both parts of the problem. The speed of the water in the hose satisfies the continuity equation A_cs * v = rate of volume flow. A_cs is the cross-sectional area of the hose, which is about 7 cm^2, so v = rate of volume flow / A_cs = .750 L/s / (7 cm^2) = .000750 m^3 / s / (.0007 m^2) = 1.1 m/s, approx.. The water enters at rest. At the point 2.50 m above the pump its velocity is 1.1 m/s. Since change in 1/2 rho v^2 + change in rho g y + change in pressure = 0 we have (1/2 rho v_2^2 - 1/2 rho v_1^2) + (rho g y_2 - rho g y_1) + (P_2 - P_1) = 0 with v_1 = 0, v_2 = 1.1 m/s, y_1 = 0, y_2 = 2.5 m and P_1 = 3 * 10^5 N/m^2. We get 1/2 * 1000 kg/m^3 * (1.1 m/s)^2 + 1000 kg/m^3 * 9.8 m/s^2 * 2.5 m + P_2 - 3 * 10^5 N/m^2 = 0 so that P_2 = 2.74 * 10^5 N/m^2, approx.. With a .500 m loss of height the quantity rho g y will change by - 5000 Pa. With the diameter increase to 4.00 cm the velocity will reduce to about .6 m/s so 1/2 rho v^2 will reduce to about 180 Pa, which is pretty much negligible. The pressure at this point will therefore be about 2.79 * 10^5 Pa. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: V_rate = 0.750 L/s * 1000 cm^3 / 1 L * (1 m)^3 / (100 cm)^3 = 0.00075 m^3/s V=v_rate/A = (0.00075 m^3/s) / (‘pi*(0.015 m)^2) = 1.06 m/s .5*’rho*v1^2 + ‘rho*g*h1 + P1 = .5*’rho*v2^2 + ‘rho*g*h2 + P2, solve for P2. Since the bottom of the hose is referenced as 0 and the water is not moving at that point, the KE and PE portion for (1) will equal zero. P2 = P1 - .5*’rho*v2^2 - ‘rho*g*h P2 = 3.00*10^5 N/m^2 - .5*1000 kg/m^3*(1.06 m/s)^2 - 1000 kg/m^3 * 9.8 m/s^2 * 2.50 m P2 = 2.75 * 10^5 N/m^2 V(in 4 cm hose) = (0.00075 m^3/s) / (‘pi*(0.02 m)^2) = 0.597 m/s With h_1 = 0, that term equals zero in Bernoulli’s equation. H_2 = -0.500 m P2 = .5*1000 kg/m^3*(1.06 m/s)^2 + 0 + 2.75 * 10^5 N/m^2 - .5*1000 kg/m^3*(0.597 m/s)^2 - 1000 kg/m^3 * 9.8 m/s^2 * (-0.500 m) = 2.80 * 10^5 N/m^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Your Self-Critique:A little guidance was needed from the given solution but this problem really helped me to start grasping this subject. Your Self-Critique Rating:3 ********************************************* Question: Gen phy: Assuming that the water in the hose is moving much more slowly than the exiting water, so that the water in the hose is essentially moving at 0 velocity, what quantity is constant between the inside of the hose and the top of the stream? what term therefore cancels out of Bernoulli's equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Velocity would be constant, with this assumption. ‘rho*g*h + P = constant. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Velocity is 0 at top and bottom; pressure at top is atmospheric, and if pressure in the hose was the same the water wouldn't experience any net force and would therefore remain in the hose ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: query gen phy problem 10.43 net force on 240m^2 roof from 35 m/s wind. What is the net force on the roof? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The height difference will be dropped from the equation. .5*’rho*v1^2* + P1 = .5*’rhov2^2 + P2 ‘dP = -.5*’rho*’dv^2 = -.5*1.29 kg/m^3 * (35 m/s)^2 = -790 N/m^2 F = P*A = 790 N/m^2 * 240 m^2 = 189,600 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** air with density around 1.29 kg/m^3 moves with one velocity above the roof and essentially of 0 velocity below the roof. Thus there is a difference between the two sides of Bernoulli's equation in the quantity 1/2 `rho v^2. At the density of air `rho g h isn't going to amount to anything significant between the inside and outside of the roof. So the difference in pressure is equal and opposite to the change in 1/2 `rho v^2. On one side v = 0, on the other v = 35 m/s, so the difference in .5 rho v^2 from inside to out is `d(.5 rho v^2) = 0.5(1.29kg/m^3)*(35m/s)^2 - 0 = 790 N/m^2. The difference in the altitude term is, as mentioned above, negligible so the difference in pressure from inside to out is `dP = - `d(.5 rho v^2) = -790 N/m^2. The associated force is 790 N/m^2 * 240 m^2 = 190,000 N, approx. ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: gen phy which term 'cancels out' of Bernoulli's equation and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: There is such a minor change in h and density has such a small value, ‘rho*g*h cancels out. When multiplied together and added or subtracted to such other large quantities, the change will not be noticed when using the proper significant figures. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** because of the small density of air and the small change in y, `rho g y exhibits practically no change. ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: `q001. Explain how to get the change in velocity from a change in pressure, given density and initial velocity, in a situation where altitude does not change. Can you tell from your expressions whether the change in velocity, for a given pressure change, is greater, less or equal when the initial velocity is greater? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Since altitude does not change, ‘rho*g*h can be excluded from the equation. Solve the equation with all P variables on one side and all .5’rho*v^2 variables on the other. Since ‘rho and g are constant, the change in P will be equal to the negative of the change in v multiplied by ‘rho and g. ‘dv = ‘sqrt(2*‘dP/(‘rho)). There should be a negative in the square root from the initial equation, but it is dropped to get the square root. If the initial velocity is greater, it will have a lesser pressure, so the final pressure should be greater if the velocity decreases, thus a positive change in pressure. Since ‘rho*g*h is constant, we are left with .5*’rho*v^2 + P = constant. If v decreases, .5*rho*v^2 decreases, which means P must increase to remain constant. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ********************************************* Question: `q002. Water is moving inside a garden hose at 4 meters / second, and it exits the nozzle at 8 meters / second. What is the change in pressure as the water moves from the end of the hose out into the surrounding air? Neglecting the effect of air resistance: How high would the stream be expected to rise if the hose was pointed straight upward? How far would the stream travel in the horizontal direction before falling back to the level of the nozzle, if directed at 45 degrees above horizontal? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Assuming the pressure in the hose to be zero, (I am not sure if this is a good assumption) you can use Bernoulli’s equation without the ‘rho*g*h and get ‘dP = .5*(1000 kg/m^3)*(4 m/s)^2 - .5 * (1000 kg/m^3)*(8 m/s)^2 = -24000 N/m^2. If the pressure is zero in the hose, increases by 24000 Pa in the nozzle, it will return to gauge pressure zero once it exits the nozzle. Now drop P from the equation. .5*’rho*v1^2 + ‘rho*g*h1 = .5*’rho*v2^2 + ‘rho*g*h2 .5*(1000 kg/m^3)*(8 m/s)^2 + 0 = 0 + 1000 kg/m^3 * 9.8 m/s^2 * h2 H2 = 3.3 m For horizontal distance 0 = 8 m/s * sin(45) * ’dt + .5*9.8 m/s^2 * ‘dt^2 ‘dt = 0, 1.16 s ‘dx = 8 m/s * cos(45) * 1.16 s = 6.57 m confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: