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.

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.