Question

A student standing on the ground next to a tall building throws a ball straight up with a speed of 39.2 m/s. If it takes the ball 4 seconds to reach the same height as the top of the building, how tall is the building?

Answer 1

Using the equation of motion:

h = ut - 0.5gt²,

g is -ve for a rising body, g ≈ 10 m/s², t = 4s, u = 39.2m/s

Note that in this problem we are neglecting the height of the student.

h = ut - 0.5gt²,

h = 39.2*4 - 0.5*10*4² Use a calculator

h = 76.8 m

The building is 76.8 m tall.