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. Assuming the ball's initial velocity was 47 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. 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?

Explanation / Answer

suppose initial velocity was v0.

in vertical :

vertical initial velocity = v0 sin47

vertical displacement = - 0.9 m

acc = - 9.81 m/s^2

using d = uy *t + ayt^2 /2

-0.9 = v0 sin46 t - 9.81t^2 /2

4.905t^2 - 0.731v0t - 0.9 = 0 ......(i)

in horizontal,

(v0cos47 )* t = horizontal distance

0.682 v0 t = 188

v0t = 275.66

putting this in eq (i)

4.905t^2 - 0.731v0t - 0.9 = 0

4.905t^2 - 0.731(275.66) - 0.9 = 0

t = 6.42 s

and v0t = 275.66

v0 = 275.66 / 6.42 = 42.91 m/s ..............Ans

time taken to travel 116 m in horizontal,

t = 116 / (42.91 cos47 ) = 3.96 sec

vertical displacement = ( 42.91 sin47)3.96 - 9.81(3.96^2 / 2)

= 47.36 m

height above fence = 47.36 - 3 = 44.36 m

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