#$&*
course Mth 277
10/16/2011 @ 2:33p.m.
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
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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:
Vy = v0ysin65.54 = 17.36m/sec
Vx = v0xcos65.54 = 7.9m/sec
T = -1/2(9.8m/s^2)t^2 + 17.36t = 0
T = 3.54sec
@& At 17.4 m/s it would take about 1.7 sec to come to rest, and an equal time to fall back to the original position. So 3.54 sec is very reasonable.
However you aren't doing vector calculus here. You need to solve problems by the methods of the course.
What is the acceleration function `a(t), what is the velocity function `v(t), and what is the resulting position function `r(t)?
What are the initial conditions?
What therefore are the values of the integration constants?*@
This part I thought I interrupted from the book rather than a way I previously knew, but it just doesn’t seem totally right to me.
‘ds = (19.07m/s)^2/9.8m/s^2 * sin(2 * 65.54)
‘ds = 21.5m
The units work out, but I don’t know why there is a 2 * the angle and why time isn’t used.
@& 19 m/s is the initial speed, the magnitude of the initial velocity.
The acceleration isn't in the direction of the initial velocity, so you wouldn't use 9.8 m/s with this speed.*@
confidence rating #$&*:erately
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Given Solution:
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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: V(t) = dr/dt * ur + r ‘dtheta/’dt * u’theta. A(t) = [d2r/dt^2 - r(d’theta/dt)^2]ur + [r*d2’theta/dt^2 + 2*drd’theta/dt]u’theta.
R = 1/1-sin(theta) R’ = (cos’theta)/1-2sin’theta + sin^2’theta
R’’ = (-sin’theta + 2sin^2’theta - sin^3’theta + 2cos^2’theta = 2sin’thetacos^2’theta) / (1 + 4sin^2’theta + sin^4’theta - 2sin’theta + sin^2’theta - 2sin^3’theta)
‘theta = t - pi/2
D’theta/dt = 1
D2’theta/dt^2 = 0
V(t) = [(cos’theta)/1-2sin’theta + sin^2’theta]ur + [1/1-sin’theta]‘dtheta
A(t) = [(-sin’theta + 2sin^2’theta - sin^3’theta + 2cos^2’theta = 2sin’thetacos^2’theta) / (1 + 4sin^2’theta + sin^4’theta - 2sin’theta + sin^2’theta - 2sin^3’theta) - 1/1-sin’theta]ur + [2*cos’theta/1-2sin’theta+sin^2’theta]u’theta.
confidence rating #$&*:confident in my technique, but not so much in my derivatives.
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Given Solution:
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@& You're doing the right things.
Do you understand how to take the derivative of `R(t) = r(t) * u_r to get that expression for V(t)? And how to take the derivative of V(t) to get A(t)?
It's good to be able to use those formulas, and to do OK in the course that's sufficient, but as an alternative to memorizing them you should ideally be able to derive them.*@
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Question: If a shotputter throws a shot from a height of 5.5t and an angle of 53 degrees with initial speed 28 ft/s. What is the horizontal distance of the throw?
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Your solution:
Vy = 28sin53 = 22.36m/s
Vx = 28cos53 = 16.85m/s
-16t^2 + 27.87t = 0
T = 1.74sec
‘dsx = 16.85m/sec * 1.74 sec = 29.32ft.
confidence rating #$&*:y
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Given Solution:
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@& That would be a good solution in physics, but for this course you need to do this in terms of vector functions.*@
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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: The velocity in the x and y will be the same due to the angle, 10.6m/sec
-16ft/sec^2*t^2 + 10.6ft/sec*t + 40ft = 0
T = 1.95sec
10.6ft/sec* t = 20.57ft….this would be the ‘dsx if the kid threw it from the edge, but we have to subtract 10 feet because he didn’t want to fall off the cliff, so our ‘dsx is 10.67feet.
confidence rating #$&*:y
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Given Solution:
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Question: **A gun is fired with muzzle speed 700ft/s at an angle of 20degrees. It overshoots the target by 60 ft. If the target is moving away from the gun at a constant speed of 15ft/s and the gunner takes 30 seconds to reload, at what angle should the second shot be fired with the same muzzle speed?
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Your solution: -16ft/sec^2 * t^2 + 239ft/sec*t = 0
T = 14.9sec
At t = 14.9sec, the target us at 9,740 feet, after reloading, the target is at 10,190feet.
This is where I start to have some questions, I know I could find the ‘dt for the bullet to make the 10,190 feet up if I fired it horizontal and I knew it would cover the distance and not drop(which is obviously not the case) I know that we need to more or less do the calculation to get it to the target at 10,190 feet, then again for that new ‘dt. After reloading, the target moved 450 more feet, so the bullet needs to be at an angle to cover the initial 9740 feet, the new 450 feet, and the extra distance that the target would travel during this new flight time of about 17 or so seconds. I think I could just guess and work my way into the answer through estimates, but I’m not 100% about the right way.
@& The trajectory of the bullet is a vector function `R_bullet(t), where we know v0 but not theta_0. The position of the car is given by another vector function `R_car(t), which you can write out from the given information.
Be sure to reference t in both functions with the same instant (e.g., with the instant of firing of the first shot, or the second if you prefer).
Set the two vector functions equal and solve for theta_0.*@
confidence rating #$&*:
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Given Solution:
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#*&!
@& My main observation is that you've got the physics down pretty well, but you aren't solving your problems in the language of vector functions.
Check my notes.*@