Question

A pitched ball is hit by a batter at a 47◦ angle. It just clears the outfield fence, 98 m away. The acceleration of gravity is 9.8 m/s 2 . Find the velocity of the ball when it left the bat. Assume the fence is the same height as the pitch. Answer in units of m/s.

Answer 1

Answer:u=31.02 m/s

Step-by-step explanation:

Given

launch angle of ball
`\theta =47^(\circ)`

Range of ball
`R=98 m`

considering ball to be a Projectile

Range of Projectile is given by

`R=(u^2\sin 2\theta )/(g)`

Where
`u=initial\ Velocity`

`\theta =launch\ angle`

`g=acceleration\ due\ to\ gravity`

`98=(u^2\sin 2(47))/(9.8)`

`98* 9.8=u^2\sin (94)`

`u^2=(960.4)/(sin (94))`

`u^2=962.74`

`u=31.02 m/s`