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.
[tex]R = u_x t[/tex]
[tex]R = u cos \theta \times t[/tex]
[tex]105 = ut \times cos 31^0[/tex]
ut = 122.5
Using
[tex]s = ut + \dfrac{1}{2}gt^2[/tex]
[tex]4 = u (sin \theta) t - \dfrac{1}{2}gt^2[/tex]
[tex]4 = u (sin 31^0) t - \dfrac{1}{2}gt^2[/tex]
[tex]4 = 122.5 \times 0.52 - 0.5\times 9.8 \times t^2[/tex]
[tex]t^2 = 12.06[/tex]
[tex]t = \sqrt{12.06}[/tex]
t = 3.47 s
now,
[tex]u = \dfrac{122.5}{3.47}[/tex]
u= 35.30 s
Speed of the ball when it leaves the bat u= 35.30 s
