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Question
A ball is thrown vertically upwards. It returns 6 s later. Calculate the greatest height reached by the ball. (Take g = 10 m s−2)
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Solution
Given:
Total time of journey = 6 s
g = 10 m/s2
Let h be the greatest height.
From the equation of motion,
`h = ut + 1/2 g t^2`
(we consider motion of ball from highest point to the ground.)
Since the total time for journey of the ball in air = 6 s.
Therefore, time for reaching maximum height = `6/2` = 3 s
For ascent, initial velocity, u = 0
Substituting the values in the formula, we get,
`h = (0 xx 3) + (1/2 xx 10 xx 3^2)`
h = 0 + (5 × 9)
h = 45 m
Hence, greatest height reached by the ball = 45 m
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