Aball is thrown straight up. it passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 1.30 s to go past the window. what was the ball's initial velocity?

u =  13.67 m/s

Explanation:

given,

window height = 2 m

window is 7.5 m off the ground on its path up

total distance from the ground to pass the window = 2 + 7.5 = 9.5 m

time taken to go past the window = 1.30 s

using equation of motion

u =  13.67 m/s

hence, the initial velocity of the  ball is equal to 13.67 m/s