physics ball

Anonymous

2013-08-13 9:59 am

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 53∘ 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?

回答 (1)

天同

2013-08-13 9:25 pm

✔ 最佳答案

Let U be the initial velocity of the ball.
Consider the horizontal motion:
188 = (U.cos(53)).t where t is the time of travel of the ball
hence, t = 188/(U.cos(53)) ------------- (1)

Consider the vertical motion:
use equation: s = ut + (1/2)at^2
with s = -0.9 m, u = U.sin(53), a = -g(=-10 m/s^2), t = 188/(U.cos(53))
hence, -0.9 = [U.sin(53)].[188/(U.cos(53))] + (1/2).(-10).[188/(U.cos(53))]^2
i.e. -0.9 = 188.tain(53) - 5.(188)^2/(U.cos(53))^2
solve for U gives U = 44.14 m/s

The initial speed of the ball is 44.14 m/s

Time of flight of the ball to the point above the fence t' is
t' = 116/[44.14.cos(53)] s = 4.367 s
Consider the vertical motion:
Use equation: s = ut + (1/2)at^2
with u = 44.14sin(53) m/s = 36.59 m/s, t = 4.367 s, a = -g(= -10 m/s^2), s =?
hence, s = [36.59 x 4.367 + (1/2)(-10).(4.367)^2] m = 64.43 m

Thus, height above fence = (64.43 + 0.9 - 3) m = 62.34 m

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