Question

Liam throws a water balloon horizontally at 8.2 m/s out of a window 18 m from the ground.

Answer 1

Time taken by the water balloon to reach the bottom will be given as

$h = \frac{1}{2} gt^2$

here we know that

$h = 18 m$

$g = 9.8 m/s^2$

now by the above formula

$18 = \frac{1}{2}*9.8* t^2$

$18 = 4.9 t^2$

$t = 1.92 s$

now in the same time interval we can say the distance moved by it will be

$d = v_x * t$

$d = 8.2 * 1.92 = 15.7 m$

so it will fall at a distance 15.7 m from its initial position