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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?

Respuesta :

olemakpadu
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.