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course Phy 241
7/14 2:04 am
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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 m/s ^ 2 * 3 m)
vf = +/-7.7 m/s
magnitude = sqrt((12 m/s) ^ 2 + (-7.7 m/s) ^ 2 )
magnitude = 14.2 m/s
arctan ( (-7.7 m/s) / (12 m/s) )
= -35 deg
360 deg + (-35 deg)
= 325 deg
confidence rating #$&*:
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Question: `q002. A projectile is given an initial velocity of 20 meters/second at an angle of 30 degrees above horizontal, at an altitude of 12 meters above a level surface. How long does it take for the projectile to reach the level surface?
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Your solution:
v0 = 20 m/sec, ds = - 12 m, theta = 30 deg
v0 vertical = 20 m/s * sin (30 deg)
v0 vertical = 10 m/s
vf = sqrt((10 m/s)^2 + 2(- 9.8 m/s^2)(- 12 m)
vf = sqrt((100 m^2/s^2) + (235.2 m^2/s^2))
vf = - sqrt(335.2 m^2/s^2)
vf = - 18.31 m/s
(10 m/s - 18.31 m/s)/ 2
8.31 m/s / 2
4.15 m/s = 4.2 m/s
ds/vAve = dt
dt = - 12 m/ (4.2 m/s)
dt = 2.86 s
confidence rating #$&*:
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Self-critique (if necessary):
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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
confidence rating #$&*:
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Self-critique (if necessary):
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Self-critique rating:
If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
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Question: `q004. A projectile is given an initial velocity of 12 meters / second in a direction 37 degrees above horizontal. It is released from a height of 2 meters and lands on a platform whose height is 3 meters. How much time elapses, and how far does it travel in the horizontal direction between release and landing?
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Your solution:
v0 = 12 m/s, ds = 3 m, theta = 30 deg
v = 12 m/s cos(30 deg)
v = 10.39 m/s
vf = sqrt((10.39 m/s)^2 + 2(9.8 m/s^2)(1 m))
vf = sqrt(107.95 m^2/s^2 + 19.6m^2/s^2)
vf = sqrt(127.55 m^2/s^2)
vf = 11.29 m/s
(12.91 m/s - 11.29 m/s)/2
= 1.62 m/s /2
= .81 m/s
dt = 1 m/ .81 m/s
dt = 1.23 s
ds = .81 m/s * 1.23 s
ds = .65 m
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Self-critique (if necessary):
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Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
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
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
(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
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If
`dt = 1 m / (.81 m/s), then it follows by simple algebraic rearrangement that
.81 m/s * `dt = 1 m.
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However the question asked how far the projectile travels in the horizontal direction. So your vertical average velocity .81 m/s will not help you answer that question.
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confidence rating #$&*:
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Self-critique (if necessary):
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Self-critique (if necessary):
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Self-critique rating:
#*&!
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