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

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

$R=\frac{u^2\sin 2\theta }{g}$

Where $u=initial\ Velocity$

$\theta =launch\ angle$

$g=acceleration\ due\ to\ gravity$

$98=\frac{u^2\sin 2(47)}{9.8}$

$98\times 9.8=u^2\sin (94)$

$u^2=\frac{960.4}{sin (94)}$

$u^2=962.74$

$u=31.02 m/s$