A ball thrown vertically upward is caught by the thrower after 25 seconds. Find the initial velocity of the ball and the maximum height it reaches.

initial velocity = u, distance = s = 0 when ball comes back, t = time = 25, acceleration = g = - 32ft/sec

s = ut + 1/2 gt^2 (By t^2 I mean square of t)

0 = 25 u - 16 25^2,

u = 400 ft/sec = Initial velocity

For the second part:

u = 400, acceleration a = -g = -32, v = final velocity = 0

Formula: v^2 -u^2 = 2as

-400^2 = -64s

s = 2500 ft.

Height = 2500 ft