A water balloon is thrown horizontally with a speed of 8.31 m/s from the roof of a building of
height H=23.0 m. How far does the balloon travel
horizontally before striking the ground? (Use the
vertical formula Av=v0v+1 2aw/2 to solve for time
and plug your components into the equation for
the horizontal direction Ar-vxt)
Your answer

Explanation:

Given

speed of water balloon = 8.31m/s

Height of building = 23.0m

Required

Horizontal distance

First we need to get the time using the equation of motion

S  = ut + 1/2at²

23 = 0+1/2(9.81)t²

23 = 4.905t²

t² = 23/4.905

t² = 4.689

t = √4.689

t = 2.17secs

Next is to get the horizontal direction expressed as:

Sh = Vxt

Sh =  8.31 * 2.17

Sh = 17.99m

Sh ≈ 18m

Hence the horizontal distance of the object before striking the ground is approximately 18m

0.11 m/s and 0.28 m/s should be correct

see earlier answer