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BlackZzzverrR [31]
3 years ago
9

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.
Physics
1 answer:
musickatia [10]3 years ago
5 0

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

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