a man is standing 12 feet above the ground on a platform. he throws a ball up in the air with an initial velocity of 20 ft/sec.

a-find the equations modeling the height of the ball and the velocity of the ball in simplest form

b-when will the ball reach its max height?

c-what is the velocity at the time the ball reaches the max height?

d-what is the velocity of the ball when it hits the ground?


Expert Answers

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The man standing on a platform at a height of 12 ft. from the ground throws a ball vertically upwards at 20 ft/s.

The velocity of the ball at any time t is given by V = 20 - 32.17*t ft/s

The height of the ball at time t is H = 20*t - (1/2)*32.17*t^2 ft.

When the ball reaches it maximum height, its velocity is 0. The time t when the ball reaches its maximum height is given by 0 = 20 - 32.17*t

=> t = 20/32.17 s.

The velocity of the ball when it returns to its 12 ft high base is 20 ft/s in a direction vertically downwards. The velocity gained in moving down 12 ft is sqrt(2*32.17*12). The velocity of the ball when it hits the ground is 20 + sqrt(2*32.17*12) = 47.78 ft/s.

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