Question

A ball is dropped from a height. If it takes 0.2s to cross the last 6m before hitting the ground, find the height from which it is dropped. Take g = 10m/s² (Ans: 54.45m)

Answer 1

For last 6 m distance travelled s = 6 m, u = ? t = 0.2 sec, a = g = 9.8 m/s2 S = ut + ½ at2 ⇒ 6 = u (0.2) + 4.9 x 0.04 ⇒ u = 5.8/0.2 = 29 m/s. For distance x, u = 0, v = 29 m/s, a = g = 9.8 m/s2 S = (V^2-u^2)/2a = (〖29〗^2 - 0^2 )/(2 x 9.8 ) = 42.05 m Total distance = 42.05 + 6 = 48.05 = 48
I hope you got it