Answer to Question #282363 in Physics for Naba

Question

A stone is thrown vertically upward it took four seconds to return to the ground how high does the stone go?

Answer 1

Let's first find the time that the stone takes to reach its maximum height:

$$t=\dfrac{t_{flight}}{2}=\dfrac{4\ s}{2}=2\ s.$$

Finally, we can find maximum height:

$$h_{max}=v_0t+\dfrac{1}{2}gt^2=0+\dfrac{1}{2}\times9.8\ \dfrac{m}{s^2}\times(2\ s)^2=19.6\ m.$$