You can use any coordinate system you like to solve aprojectile motion problem. To demonstrate the truth of this statement, consider a ball thrown off the top of a building witha velocity vS at an angle u with respect to the horizontal. Letthe building be 50.0 m tall, the initial horizontal velocity be9.00 m/s, and the initial vertical velocity be 12.0 m/s. Chooseyour coordinates such that the positive y - axis is upward, thex - axis is to the right, and the origin is at the point where theball is released. (a) With these choices, find the ball’s maximumheight above the ground and the time it takes to reachthe maximum height. (b) Repeat your calculations choosingthe origin at the base of the building.

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Asked Dec 17, 2019
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You can use any coordinate system you like to solve a
projectile motion problem. To demonstrate the truth of this statement, consider a ball thrown off the top of a building with
a velocity vS at an angle u with respect to the horizontal. Let
the building be 50.0 m tall, the initial horizontal velocity be
9.00 m/s, and the initial vertical velocity be 12.0 m/s. Choose
your coordinates such that the positive y - axis is upward, the
x - axis is to the right, and the origin is at the point where the
ball is released. (a) With these choices, find the ball’s maximum
height above the ground and the time it takes to reach
the maximum height. (b) Repeat your calculations choosing
the origin at the base of the building.

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Expert Answer

Step 1

Given:

Height of the building = 50 m

Initial horizontal velocity = 9 m/s

Initial vertical velocity = 12 m/s

Step 2

Resultant velocity can be calculated as

 

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u² +u, =u 9² +12² =u u =15m/s but, horizontal velocity = 9 m/s ucos 0 = 9 15cos 0 = 9 cos 0 = 15 0 = 53.13°

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Step 3

Part (a):

Let, maximum height covered by the ball = h

At maximum hei...

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v - u = -2gh 0-144 = -2x9.8×h 144 h= = 7.35m 19.6

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