Please complete the problems showing all calculations and equations on which you

Question

Please complete the problems showing all calculations and equations on which you base your answers. Use -9.8 m/s2 as the value for the accelarattion due to gravity.

A baseball is struck by a bat 48 cm from the bat's axis of rotation when she bat's angular velocitity is 4000 deg/sec. Assume that the ball leaves bat with the same linear velocity as the contact point of the bat. If the ball is hit at a height 1.0 m above the ground at a 50 degree angle above the horizontal, with the ball clear the outfiield fence which is 2 m. in height, at a horizaontal distance 108 m. form the point of ball contact with the ball? (hint: this is multi part problem)

Explanation / Answer

angular velocity of the bat, w = 4000 deg/s

= (4000/(360)) rad/s

= 11.11 rev/s

= 11*11*2*pi rad/s

= 69.8 rad/s

linear speed at contact point, vo = r*w

= 0.48*69.8

= 33.5 m/s

vox = vo*cos(50) = 33.5*cos(50) = 21.5 m/s

voy = vo*sin(50) = 33.5*sin(50) = 25.66 m/s

vertical displacement needed, h = 2 - 1 = 1 m

let t is the time taken.

Apply, h = voy*t + 0.5*g*t^2

1 = 25.66*t - 4.9*t^2

4.9*t^2 - 25.66*t + 1 = 0

on solving the above equation we get

t = 5.197 s

noe Apply horizontal distance travelled when the height beomces 2m above the ground,

x = vox*t

= 21.5*5.197

= 111.7 m

celary 111.7 m > 108 m

so, the ball can clear the fence.

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