Question

A student drops a ball from the top of a 37-meter tall building. The ball leaves the student's hand with a zero speed. What is the speed of the ball at the moment just before it hits the ground?

Answer 1

The ball accelerates because of gravity which is constant. So we can apply one of the equations of uniformly accelerated motion to solve the problem.

`Vf^2=Vo^2+2aS\\` is compatible since
`S, Vo, a` are given enabling us to calculate
`Vf` .

`Vf^2=(0m/s)^2+2*10m/s^2*37m\\Vf^2=20m/s^2*37m\\Vf^2=740m^2/s^2\\Vf=√(740) m/s\\ Vf=27.202941017 m/s\\`