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course Phy 241
08/6 6:28 pm
Question: `q001. Note that this assignment contains 4 questions.
. A projectile has an initial velocity of 12 meters/second, entirely in the horizontal direction. At the instant of first contact with a level floor three meters lower than the initial position, what will be the magnitude and direction of the projectile's velocity vector?
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Your solution:
vf = +-`sqrt( 0^2 + 2 * 9.8 meters/second ^ 2 * 3 meters)
vf = +/-7.7 meters/second
magnitude = sqrt((12 m/s) ^ 2 + (-7.7 m/s) ^ 2 )
magnitude = 14.2 m/s
arctan ( (-7.7 meters/second) / (12 meters/second) )
= -35 degrees
360 degrees + (-35 degrees)
= 325 degrees
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Question: `q002. A projectile is given an initial velocity of 20 meters/second at an angle of 30 degrees above horizontal, at an altitude of 12 meters above a level surface. How long does it take for the projectile to reach the level surface?
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Your solution:
v0 = 20 m/sec, ds = - 12 m, theta = 30 deg
v0 vertical = 20 m/s * sine (30 deg)
v0 vertical = 10 m/s
vf = sqrt((10 m/s)^2 + 2(- 9.8 m/s^2)(- 12 m)
vf = sqrt((100 m^2/s^2) + (235.2 m^2/s^2))
vf = - sqrt(335.2 m^2/s^2)
vf = - 18.31 m/s
(10 m/s - 18.31 m/s)/ 2
8.31 m/s / 2
4.15 m/s = 4.2 m/s
ds/vAve = dt
dt = - 12 m/ (4.2 m/s)
dt = 2.86 s
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Question: `q003. What will be the horizontal distance traveled by the projectile in the preceding exercise, from the initial instant to the instant the projectile strikes the flat surface.
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Your solution:
dt = 2.8 s, v0 = 20 m/s, theta = 30 degree
v horizontal = 20 m/s*cos(30 deg)
v horizontal = 17.32 m/s
ds = 17.32 m/s * 2.8 s
ds = 48.5 m
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If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
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Question: `q004. A projectile is given an initial velocity of 12 meters / second in a direction 37 degrees above horizontal. It is released from a height of 2 meters and lands on a platform whose height is 3 meters. How much time elapses, and how far does it travel in the horizontal direction between release and landing?
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Your solution:
v0 = 12 m/s, ds = 3 m, theta = 30 deg
v = 12 m/s cos(30 deg)
v = 10.39 m/s
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If this is to be the initial vertical velocity, which I assume is the case because you're using it with the 9.8 m/s^2 acceleration of gravity, then you wouldn't use the cosine to find this component.
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vf = sqrt((10.39 m/s)^2 + 2(9.8 m/s^2)(1 m))
vf = sqrt(107.95 m^2/s^2 + 19.6m^2/s^2)
vf = sqrt(127.55 m^2/s^2)
vf = 11.29 m/s
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The solution to the preceding equation is +- 11.29 m/s.
The projectile wouldn't land on the platform with a positive vertical velocity.
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(12.91 m/s - 11.29 m/s)/2
= 1.62 m/s /2
= .81 m/s
dt = 1 m/ .81 m/s
dt = 1.23 s
ds = .81 m/s * 1.23 s
ds = .65 m
confidence rating #$&*:
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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 responses. See my notes and let me know if you have questions. &#