Question

in a tall building, a certain window is 500 meters above the ground.A tennis ball is catapulted horizontally out of the window with a horizontal initial velocity of 4 m/s. How far from the base of the building will the ball land?

Answer 1

Horizontal speed of the ball is given as

`v_x = 4 m/s`

height of the ball is given as

`h = 500 m`

now the time taken by the ball to reach the bottom is given as

`h = v_i*t + (1)/(2)gt^2`

now we will plug in all values

`500 = 0 + (1)/(2)*9.8*t^2`

`t = 10.1 s`

now the distance from the base is given as

`x = v_x * t`

`
`x = 4 * 10.1 = 40.4 m`

so it will land at distance of 40.4 m