Math
posted by Candice on .
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 :)

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 = (6552) / 1.857 = 7.0 m/s. =
Speed of fieldsman.