#$&*
course Mth 277
10/7 4
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
*********************************************
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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
using the given equations in the book....
fill in for v0 and alpha and gravity to get the answers..
Tf = 2/g v0sin alpha = 3.5 seconds
Rf = v0^2/g sin 2alpha = 27.97 meters
@& You need to go through the process. Otherwise you won't know what to do if the conditions vary.
Integrating a = -9.8 `j you get
v(t) = c1 `i + (-9.8 t + c2) `j
v(0) = (v0 cos(alpha) ) `i + (v0 sin(alpha) ) `j) so
v(t) = (v0 cos(alpha) )`i + (v0 sin(alpha) - 9.8 t ) `j.
Integrate v(t) to get
r(t) = (v0 cos(alpha) * t + c1) `i + (v0 sin(alpha) t - 4.9 t^2 + c2) `j.
Assume that r(0) = 0. Then c1 and c2 are both zero. You need to find t such that the `j component is zero, since the projectile returns to the ground.
So you solve
v0 sin(alpha) t - 4.9 t^2 = 0
which is
19.07 sin(65.54 deg) - 4.9 t^2 = 0
and find that
t = 0 or t = 19.07 sin(65.54 deg) / 4.9.
This comes out to about 3.5 seconds.
The x component is found by substituting this value of t into the expression for the x component, which does come out around 30 meters.
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
*********************************************
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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
after messy calculations, use the formula V(t) = (dr/dt)*u_r + r(dr/dt)*u_theta
combine and eliminate some terms to get an equation for V(t) = -2sin(theta) + sin(theta)cos(theta) + sin^2(theta) / ( 1-2sin(theta) + sin^2(theta) )
fill in (1-pi/2) for theta and get a velocity of .387 rad/s
and to get A(t) just take the derivative of the V(t) to get;
A(t) = -2cos(theta) + ( -sin^2(theta) + cos^2(theta) ) + sin2(theta) / ( -2sin(theta)cos(theta) )
fill in for theta = t-pi/2 to get -2.4 rad/s^2 this implies that the object is slowing down
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I'm not sure about most of this. this is very messy
------------------------------------------------
Self-critique rating:
@& dr/dt = dr/dTheta* dTheta / dt.
dTheta / dt = 1 so
dr/dt = cos(theta) / (1 - sin(theta))^2 * 1 = cos(t - pi/2) / (1 - sin(t - pi/2))^2
r dTheta / dt = r = 1 / (1 - sin(t - pi/2)).
So
v(t) = cos(t - pi/2) / (1 - sin(t - pi/2))^2 * `u_r + r dTheta / dt = r = 1 / (1 - sin(t - pi/2)) ( `u_theta.
a(t) also ends up with both u_r and u_theta terms.*@
@& It does get messy, but less so if you keep track of the coefficients of `u_r and `u_theta.*@
*********************************************
Question: If a shotputter throws a shot from a height of 5.5t and an angle of 53 degrees with initial speed 28 ft/s. What is the horizontal distance of the throw?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
first find the Tf = 2/32.2ft/s^2 (28ft/s)sin53 = 1.4 seconds. now fill in the seconds to the x(t) = (V_0cos alpha) t
to get 28ft/s cos(53)(1.4s) = 23.6ft
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
OK
------------------------------------------------
Self-critique rating:
@& This doesn't take account of the initial vertical position being 5.5 ft.
You need to integrate the acceleration function twice. a = -32.2 `j, and go from there.
Your intial condition on velocity is v(0) = 28 ft / s * cos(53 deg) * `i + 28 ft / s * sin(53 deg) * `j.
The simplest initial position is 5.5 `j.
You need to find t such that the `j component of the position function is 0.*@
*********************************************
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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
(somehow i think this question is based on your grandson)
assuming v_0 is 15 ft/s and not counting any added thrust from the childs arm.....
we know to find this we need to have the t value for x(t) = 15ft/s cos45 t. so we solve a parametric equation for t in the y(t) equation;
y(t) = -1/2(32ft/s^2) t^2 (15ft/s sin45) t + 40ft.
we are left with -16t^2 +10.6t +40.
now because when y(0), the rock will be on the ground, or in this case, the water, so set y(0) and get; t = 1.84s
fill in x(1.84) = 15ft/s cos45 t. to get 19.5 ft.
now because he threw the rock 10 ft from the edge, we subtract 19.5 ft - 10ft = 9.5 ft.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
seems a short distance, but i guess 15 ft/s is just a lob rather than a full on throw
------------------------------------------------
Self-critique rating:
@& Good. Can't confirm the numbers precisely but they're plausible, and your equations are correct.
The x(t) and y(t) equations should come from integrating a = -32 `j and applying the initial conditions.
15 ft / s isn't much. That would be barely higher than the ceiling.*@
*********************************************
Question: **A gun is fired with muzzle speed 700ft/s at an angle of 20degrees. It overshoots the target by 60 ft. If the target is moving away from the gun at a constant speed of 15ft/s and the gunner takes 30 seconds to reload,
at what angle should the second shot be fired with the same muzzle speed?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
This is cool, but very difficult
I can solve for t when it hits the target for t = 14.97 seconds. round to 15s.
the target is 450ft away after 30seconds,
we set the equation y(t) = -1/2(32.2ft/s^2)(15s)^2 + (700ft/s sin(theta))(15s) +0 = 450ft
3,600 ft + 450 ft / 700ft/s = 5.78 s/ft then / 15s = .3857. take arc sin (.3857) to get angle of 22.69 degrees.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
That answer should be very close to right, but i know my units are iffy on that last line of operations.
------------------------------------------------
Self-critique rating:
@& Not bad, but it takes the second shot time to get there so the target will be further away than that.
In 15 seconds the bullet will travel close to 10 000 feet, not 3600 ft.
The target's position will be this distance, minus 60 feet, plus 15 ft / s * t, where t is measured by a clock that starts at the instant the first bullet hits the ground.
The bullet's speed in the horizontal direction is 700 ft / sec * cos(theta), where theta is the angle of elevation.*@
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!
@& Pretty good, but as you might expect some of the problems are more complicated (and some maybe a little less) than you expect.
Check my notes.*@