According to the Guinness Book of World Records, the longest home run ever measu

Question

According to the Guinness Book of World Records, the longest home run ever measured was hit by Roy Dizzy Carlyle in a minor league game. The ball traveled 188 m (618 ft) before landing on the ground outside the ballpark.

A-

Assuming the ball's initial velocity was 54?above the horizontal and ignoring air resistance, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point 0.9 m (3.0 ft) above ground level? Assume that the ground was perfectly flat.

Express your answer using two significant figures.

B-

How far would the ball be above a fence 3.0 m (10 ft) high if the fence was 116 m(380 ft) from home plate?

Express your answer using two significant figures.

Explanation / Answer

We have to determine the flight time. Ignoring air resistance, the time from the launch height to max height is equal to the time from max height back to launch height. At max height, Vy(t) = 0. Vy slows to a stop before falling at g. Vy(t) = Vyo - gt => Vyo/g = t where t is the time to max height from launch height The time to fall back to the launch height is t = Vyo/g At the moment the ball reaches the launch height Vxf=Vxo and Vyo = -Vyo At h=0.9m the ball has not traveled 188m. We we can draw a right triangle with height 0.9, angle 56 degrees because ignoring air resistance, the impact angle is 56 degrees from the negative x axis. Therefore, the difference in the horizontal distance at the launch height is x and 0.9/x = tan56 so x = 0.9/tan56 = 0.607m 188-0.607=187.4m which is the distance traveled in 2t = 2Vyo/g at Vxo tan56=Vyo/Vxo => Vyo = Vxotan56 Vxo*2t = Vxo*2*Vxo*tan56/g = 187.4 => sqrt(187.4*g/2*tan56) = Vxo=24.9m/s Vyo=36.9 and Vo = 44.5m/s at d = Vxo*t = 116 => t = 116/Vxo = 4.66s h(t) = 0.9 + 44.5*sin56 * 4.66 - 4.9*4.66^2 = 66.4m

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