Question

A ball 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?

Answer 1

u = 13.67 m/s

Step-by-step 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

`S = u t +(1)/(2)gt^2`

`(2+7.5)= u* 1.3 -0.5* 9.8* 1.3^2`

`u* 1.3 = 9.5 + 8.281`

u = 13.67 m/s

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