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

$h = v_i*t + \frac{1}{2}gt^2$

now we will plug in all values

$500 = 0 + \frac{1}{2}*9.8*t^2$

$t = 10.1 s$

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