Query103

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course Mth 277

530am 11/13/2011

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.

query_10_3

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Question: Find the time of flight Tf (to the nearest tenth of a second) and the range Rf (to the nearest unit) of a projectile fired

(in a vaccum) from ground level at `alpha = 65.54 degrees and v0 = 19.07 m/s. Assume that g = 9.8 m/s^2.

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Your solution:

r(t) = (v_0c*cos(theta)t i + [h + (v_0*sin(theta)t - .5gt^2] j

assume ground level is h=0

r(t) = 19.07 m/s * cos(65.54)t i + [0 + 19.07 m/s * sin(65.54)t - .5(9.8 m/s)t^2]

= (7.9 m/s)t i + [(17.4 m/s)t - (4.9 m/s)t^2] j

If projectile is shot from h=0 and lands at h=0. So in order to find the time it takes to land, we set h=0 and solve the j component, which is a parabala and we can find the zeros of this parabola

So,

[(17.4 m/s)t - (4.9 m/s)t^2]

Use the quadratic formula to find the zeros

t= 0, 3.6 s

The range of the projectile is found by solving the i component when t= 3.6 s

range = (7.9 m/s)t

= 28.44 m

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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Question: An object is moving along the curve r = 1/(1 - sin(theta)), theta = t - pi/2.

Find its velocity and acceleration in terms of the unit polar vectors u_r and u_theta.

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Your solution:

R'(t) = r'(t)* u_r + r(t) * d(theta)/'dt * u_theta

R""(t) = (r"" - r * [d(theta)/'dt]^2 * u_r + [2r'*d(theta)/'dt + r*d^2(theta)/'dt^2] *u_theta

r= 1/ (1-sin(theta)

theta = t- pi/2

If I were to complete the solution I would find the first and second derivatives for r and theta. and plug and chug. But after hours spent on this one problem, I am still having difficulty.????????????????/

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d(theta) / dt = (d ( t - pi/2) ) / dt = 1.

d^2(theta) / dt = 0.

r ' = dr / dt = 1 / (1 - sin( t - pi/) ) = etc..

(can also be calculated as dr/dtheta * dtheta / dt)

r '' can be calculated from r ' .

It should be straightforward, after calculating these derivatives, to obtain and to a reasonable extent simplify the expressions for r ' and r ''.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I had a really hard time with this one. I wasn't sure whether to stay in polar coordinates or convert to rectangular. Then I thought, well, I will just find R(t) = rcos(theta)i + rsin(theta)j

and find the derivative of that to find velocity function and find the derivative of that to find the acceleration function. So, once I started doing that, I wasn't sure whether to plug in r and r' and r'' in later once I had solved for the first and second derivatives of the R(t)

So I went back to the videos of Chapter 10 and watched 22 thru 28 over and over again and I couldn't grasp the idea of U_theta and u_r. I understand that they are unit vectors of the position function broken down into components. I understand that velocity is tangent to the position vector. I want to say that the acceleration has a y component that is tangent to the position vector and x component normal to the tangent vector??????????????? I'm not sure about that though. I am on the cusp of understanding the concepts. I know u_theta = -sin(theta)i + cos(theta)j and u_r = cos(theta)i + sin(theta)j and I understand why.

I'm just having a hard time combining all of the 'new' concepts. Though polor coordinates should not be 'new' to me, and I feel that after going through the videos for chapter 10 and 11 I am pretty well versed in the r and theta, but the polar coordinates are still alittle foriegn to me as opposed to the rectangular coordinates, which is understandable.

???????????????

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Self-critique rating:

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Question: If a shotputter throws a shot from a height of 5.5t and an angle of 53 degrees with initial speed 28 ft/s. What is the horizontal distance of the throw?

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Your solution:

Use the position function for a projectile:

h = 5.5t

theta = 53°

v_0 = 28 ft/s

g= 32 ft/s^2

r(t) = (v_0cos(theta)t i + [h + v_0sin(theta)t - .5gt^2] j

Find the value of t, when the j component is equal to zero, assuming the height of the shotputter is 5.5 and the height of the ground is 0

5.5*t + 28ft/s*sin(53)t - 16ft/s^2*t^2 = 0

Use the quadratic formula to find the zeros of this equation

t= 0 and approximately 5 seconds when the j component is equal to zero

Plug t=5 s into the i component of the equation

28ft/s*cos(53)*(5s)

= approximately 85.4 ft

confidence rating #$&*:

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Nobody can throw the shot quite that far (I think world records are in the 70's). However the error is in the statement of the problem. 5.5 t should read 5.5 ft.

The typo results in an additional vertical velocity, resulting in greater time in the air.

With the given information your solution is correct.

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Given Solution:

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Self-critique (if necessary):

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Question: A child running along level ground at the top of a 40ft high vertical cliff at a speed of 15ft/s, throws a rock over the cliff into the sea below.

If the rock is released 10 ft from the edge and at an angle of 45degrees, how long does it take the rock to hit the water and how far away from the base of the

cliff does it hit?

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Your solution:

v_0 = 15 ft/s

h= 40 ft

r(t) = (v_0cos(theta)t i + [h + v_0sin(theta)t - .5gt^2] j

v(t) = -v_0sin(theta)i + [v_0cos(theta) - gt]j

a(t) = -v_0cos(theta)i + [-v_0sin(theta) - g]j

r(t) = (v_0cos(theta)t i + [h + v_0sin(theta)t - .5gt^2] j

= [15 ft/s * cos(45)t] i + [40 + 15 ft/s*sin(45)t - 16ft/s^2*t^2] j

Set the x-axis at the sea-level, so h = 40 ft and the j component for the position vector must be equal to zero to find t, when the rock hits the water.

[40 + 15 ft/s*sin(45)t - 16ft/s^2*t^2] = 0

Use the quadratic formula to find the zeros of the function

t= 9.96 and 11.25 seconds

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Right method, but using t = 10 sec yields a height of 40 ft + 100 ft - 1600 ft = -1460 ft, not 0. So your solution to the quadratic would need to be revised (easy enough to do).

Otherwise good solution.

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plug both values into the i component of the position vector

(v_0cos(theta)t = [15 ft/s * cos(45)t]

= 15 ft/s * cos(45)*9.96 s

= 105.64 ft

and 15 ft/s * cos(45)*11.25 s

= 119.32 ft

I would go with the lowest values t= 10 s and s = 106 ft because this is the first intercept with the x-axis whish I have assumed is sea level.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question: **A gun is fired with muzzle speed 700ft/s at an angle of 20degrees.

It overshoots the target by 60 ft. If the target is moving away from the gun at a constant speed of 15ft/s and the gunner takes 30 seconds to reload,

at what angle should the second shot be fired with the same muzzle speed?

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Your solution:

I used the postion function for a projectile

r(t) = (v_0cos(theta)*t i + (h + v_0sin(theta)*t -.5gt^2) j

v_0 = 700 ft/s

theta = 20°

h=0

g= -32 ft/s^2

In order to find the time it took for the bullet to overshoot the target I set the j component to zero, because the x-axis is the ground.

[700ft/sin(20°)*t - .5(-32ft/^2)*t^2] j

I used the quadratic formula to find t

t= 0, 14.96s

In approximately 15 s, the bullet overshot the target by 60 ft. Common sense tells me 15 seconds for a bullet to land is hard to imagine, but 20 degrees is not a very big incline.

So I plug t=15 seconds into the x component of the postion function.

700ft/s*cos(theta)*15s + 390 ft

I found 390 ft by subtracting 60 ft, which was the overshoot, from 450ft which is the adjusted target range, because it take 30 seconds at 15ft/s for the target to move 450 ft.

So in 15 seconds the bullet can travel horizontally 10,950 feet. nearly two miles! That sounds alittle extreme. But either way, I set my i component equal to this value.

700ft/s*cos(theta)*15s + 390 ft = 10,950 ft.

theta = 61°

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

15 seconds for a bullet to travel and 2 miles is alittle extreme.

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Self-critique rating:

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The vertical component of the velocity will be over 250 ft / sec; to slow to zero would take over 7 seconds, and it would require another 7 seconds (plus) to come down. So 14.96 s is not unreasonable.

The bullet requires additional time to go 450 ft. so if the adjustment simply aims 450 ft further, it will fall short of the target by several feet. You have to treat time as a variable.

Still, not a bad attempt.

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

#*&!

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Good work overall, and good questions. Check my notes.

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