You throw a ball straight up and catch it 4S later Determine

You throw a ball straight up and catch it 4S later. Determine the initial velocity of the ball? Find the maximum height of the ball. Calculate the velocity of the ball when it is at half the maximum height.

Solution

A. time to catch back the ball T = 4 sec

it means ball reached at maximum height at the time

t = T/2 = 4/2 = 2 sec

u = initial velocity = ?

v = 0 at max height

a = -g = -9.81 m/sec^2

v = u + a*t

0 = u - 9.81*2

u = 19.62 m/sec

B.

max height will be

H = u*t + 0.5*a*t^2

H = 19.62*2 - 0.5*9.81*2^2

H = 19.62 m

C.

half of max height = 19.62/2 = 9.81 m

v^2 = u^2 + 2*a*s

v^2 = 19.62^2 - 2*9.81*9.81 = 192.47

v = sqrt (192.47)

v = 13.87 m/sec