Question

A person standing on a 20m tall building throws a ball straight up. The ball is observed to go 5m above the point where she let it go. It then falls down to the ground.

1. How fast was the ball thrown?

2.How long did it take for the ball to hit the ground?

3. how fast is the ball going just before it hits the ground?

Answer 1

let

H = 20 m

h = 5 m

1) use, h = vo^2/(2*g)

==> vo = sqrt(2*g*h)

= sqrt(2*9.8*5)

= 9.90 m/s <<<<<<<--------Answer

2) let t is the time taken

use, y = voy*t + (1/2)*ay*t^2

-H = vo*t + (1/2)*(-g)*t^2

-20 = 9.90*t - (1/2)*9.8*t^2

on solving the above equation we get

t = 3.27 s <<<<<<<--------Answer

3) speed of the ball before hitting the ground, v = sqrt(2*g*(H+h))

= sqrt(2*9.8*(20 + 5))

= 22.1 m/s <<<<<<<--------Answer