Question

A batsman hits a cricket ball 'off his toes' towards a fieldsman who is 65 m away. The ball reaches a maximum height of 4.9 m and the horizontal compoenent of its velocity is 28 m/s. Find the constant speed with which the fieldsman must run forward, starting at the instant the ball is hit, in order to catch the ball at a height of 1.3m above the ground. (Use g = 9.8)


Thanks in advance :)

Answer 1

h = ho - 0.5g*t^2 = 0.
4.9 - 4.9*t^2 = 0
-4.9t^2 = -4.9
t^2 = 1
Tf = 1 s. = Fall time or time to
fall to Gnd.

Tr = Tf = 1 s. = Rise time or time to
rise to max. Ht.

h = ho - 0.5g*t^2 = 1.3 m.
4.9 - 4.9*t^2 = 1.3
-4.9t^2 = 1.3 -4.9 = -3.6
t^2 = 0.735
Tf' = 0.857 s. = Fall time or time to fall to 1.3 m.

Dx = Xo * (Tr+Tf').
Dx = 28m/s * 1.857s = 52 m. = Hor. dist.
traveled by the ball.

d = V*T.
V = d/T = (65-52) / 1.857 = 7.0 m/s. =
Speed of fieldsman.