# A metal ball is dropped from the roof of a building 75m high. Neglecting air resistance, what is the velocity of the ball five seconds before it reaches the ground?

## Answer

V^2 = u^2 + 2as can't be used because s is unknown.

How about v = u +at ?

t is unknown, but if you could find t when the ball hits the ground, you could just subtract 5 seconds from it and use it in the above equation.

So use s = ut + 1/2at^2

u = 0

a = g = 9.81 m/s^2

s = 75 m

So

s = ut + 1/2at^2

But u = 0

So

s = 1/2at^2

and

t = t = square root(2h/g)

Substituting

t = t = square root(2(75)/9.81) = 3.91 seconds

So 5 seconds before the ball hits the ground, the velocity of the ball is zero because it hasn't been released!

For more info on projectile motion and the equations for objects dropped, thrown up or projected at an angle from the ground, see my other tutorial:

https://owlcation.com/stem/Solving-Projectile-Moti...

Updated on March 22, 2018

### Original Article:

By Eugene Brennan