Question

A 10 pesos is dropped from the top the building. Calculate the position and the velocity of the coin after 1 second.

Answer 1

The distance traveled by coin

$$h=\frac{gt^2}{2}=\frac{9.8\:\rm{m/s^2}\times (1\:\rm{s})^2}{2}=4.9\:\rm{m}$$

So, position of the coin at the moment 1s after it was released is 5.1 meter above a ground.

The velocity of the coin

$$v=gt=9.8\:\rm{m/s^2}\times 1\:\rm{s}=9.8\:\rm{m/s}$$