Question

During a baseball game, a batter hits a pop-up to a fielder 93 m away.The acceleration of gravity is 9.8 m/s2.If the ball remains in the air for 5.4 s, howhigh does it rise?Answer in units of m

Answer 1

Explanation:

It is given that, a batter hits a pop-up to a fielder 93 m away, range of the projectile, R = 93 m

The ball remains in the air for 5.4 s, the time of flight is 5.4 s

Time of flight : $T=\dfrac{2v\sin\theta}{g}$

$5.4=\dfrac{2v\sin\theta}{g}\\\\v\sin\theta=\dfrac{5.4\times 9.8}{2}\\\\v\sin\theta=26.46$

Maximum height of the projectile : $H=\dfrac{v^2\sin^2\theta}{2g}$

We need to find H.

So,

$H=\dfrac{(v\sin\theta)^2}{2g}\\\\H=\dfrac{(26.46)^2}{2\times 9.8}\\\\H=35.72\ m$

So, it will rise to a height of 35.72 m.