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

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.