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?

Horizontal speed of the ball is given as

[tex]v_x = 4 m/s[/tex]

height of the ball is given as

[tex]h = 500 m[/tex]

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

[tex]h = v_i*t + \frac{1}{2}gt^2[/tex]

now we will plug in all values

[tex]500 = 0 + \frac{1}{2}*9.8*t^2[/tex]

[tex]t = 10.1 s[/tex]

now the distance from the base is given as

[tex]x = v_x * t[/tex]

`[tex]x = 4 * 10.1 = 40.4 m[/tex]

so it will land at distance of 40.4 m