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

u= 35.30 s

Step-by-step 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 * t`

`105 = ut * cos 31^0`

ut = 122.5

Using

`s = ut + (1)/(2)gt^2`

`4 = u (sin \theta) t - (1)/(2)gt^2`

`4 = u (sin 31^0) t - (1)/(2)gt^2`

`4 = 122.5 * 0.52 - 0.5* 9.8 * t^2`

`t^2 = 12.06`

`t = √(12.06)`

t = 3.47 s

now,

`u = (122.5)/(3.47)`

u= 35.30 s

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