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

Step-by-step 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=(2v\sin\theta)/(g)`

`5.4=(2v\sin\theta)/(g)\\\\v\sin\theta=(5.4* 9.8)/(2)\\\\v\sin\theta=26.46`

Maximum height of the projectile :
`H=(v^2\sin^2\theta)/(2g)`

We need to find H.

So,

`H=((v\sin\theta)^2)/(2g)\\\\H=((26.46)^2)/(2* 9.8)\\\\H=35.72\ m`

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