Question

A penny is dropped from the top of a building that is 300.0 m tall. Calculate the speed of the penny as it hits the ground. (meters per second please)

Answer 1

We have the equation of motion $s = ut + \frac{1}{2} at^2$, where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.

Here s = 300 m, u = 0 m/s, a = 9.81 $m/s^2$

Substituting

   $300 = 0*t+\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2 = 300\\ \\ t =7.82 seconds$

Now we have v = u+at, where v is the final velocity

Here u = 0 m/s, a= 9.81 $m/s^2$ and t = 7.82 seconds

Substituting

     v = 0+9.8*7.82 = 76.68 m/s

The speed with which the penny strikes the ground = 76.68 m/s.