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\(^{-2}\)]
(i) time taken to reach the maximum height,
(ii) speed of the throw. [g = 10 ms\(^{-2}\)]
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Correct Answer: Option n
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
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