During a baseball game, a batter hits a high pop-up. The ball remains in the air for 6.0 seconds. How high does it rise?

Question

jiteshmeghwal9 · Answer

Let $s_y$ be the displacement of ball along y-direction which is 0 as the ball came back to its initial position.
And $u_y$ be the initial velocity of ball.

The ball is moving under constant force of gravity so we may apply equations of motion here.

jiteshmeghwal9 · Answer

$$s_y=u_yt-\frac{1}{2} gt^2$$ i assumed downward direction negative and upward direction to be positive.

jiteshmeghwal9 · Answer

$$0=6u_y-\frac{1}{2} g(6)^2$$$$u_y=30 $$

jiteshmeghwal9 · Answer

Now we found the initial velocity of ball . The ball reached its maximum height when the velocity of ball becomes zero. $$v_y^2=u_y^2-2gh_{max}$$ $v_y=0$ so $$0=(30)^2-2gh_{max}$$ find $h_{max}$ which is maximum height achieved.