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?
1 Answer | Add Yours
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.
We’ve answered 319,189 questions. We can answer yours, too.Ask a question