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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.

Respuesta :

Answer:u=31.02 m/s

Explanation:

Given

launch angle of ball [tex]\theta =47^{\circ}[/tex]

Range of ball [tex]R=98 m[/tex]

considering ball to be a Projectile

Range of Projectile is given by

[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]

Where [tex]u=initial\ Velocity[/tex]

[tex]\theta =launch\ angle[/tex]

[tex]g=acceleration\ due\ to\ gravity[/tex]

[tex]98=\frac{u^2\sin 2(47)}{9.8}[/tex]

[tex]98\times 9.8=u^2\sin (94)[/tex]

[tex]u^2=\frac{960.4}{sin (94)}[/tex]

[tex]u^2=962.74[/tex]

[tex]u=31.02 m/s[/tex]

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