Question

A home run just clears a fence 105 m from home plate. The fence is 4.00 m higher than the height at which the batter struck the ball, and the ball left the bat at a 31.0° angle above the horizontal. At what speed did the ball leave the bat?

Answer 1

Answer:

u= 35.30 s

Explanation:

given,

horizontal distance covered by the ball = 105 m

vertical distance to clear by the ball = 4 m

angle at which the ball was hit = 31°

Let the initial velocity is u m/s and the ball take t sec to reach the fence.

$R = u_x t$

$R = u cos \theta \times t$

$105 = ut \times cos 31^0$

ut = 122.5

Using

$s = ut + \dfrac{1}{2}gt^2$

$4 = u (sin \theta) t - \dfrac{1}{2}gt^2$

$4 = u (sin 31^0) t - \dfrac{1}{2}gt^2$

$4 = 122.5 \times 0.52 - 0.5\times 9.8 \times t^2$

$t^2 = 12.06$

$t = \sqrt{12.06}$

t = 3.47 s

now,

$u = \dfrac{122.5}{3.47}$

u= 35.30 s

Speed of the ball when it leaves the bat u= 35.30 s