A ball thrown vertically upward reaches a maximum height of 50 m above the level of projection. Calculate the;
(i) time taken to reach the maximum height,
(ii) speed of the throw. [g = 10 ms(^{-2})]
Explanation
V(_1) = U – gt
U = V + gt = 0 + 10t
h = ut(frac{1}{2}) – gt(^2),
50 = 10t(^2) (frac{1}{2}) x 10 t(^2)
50 = 10t(^2) – 5t(^2)
50 = 5t(^2) = 10
t(^2) = 10
t = (sqrt{10}) = 3.16s
(ii) V = U – gt, O = U – 10 x 3.16
U = 31.16