Question

A ball is thrown from the top of a building with a horizontal velocity of 11.9 m/s and takes 3.54 s to hit the ground. How far above the ground, in m, is the ball after 1.47 s of falling?

Answer 1

The total height:

$$H=\frac{gT^2}{2}$$

The height of the ball after 1.47 s of falling:

$$h=H-\frac{gt^2}{2}=\frac{gT^2}{2}-\frac{gt^2}{2}$$

$$h=0.5g(T^2-t^2)=0.5(9.8)(3.54^2-1.47^2)=50.8\ m$$