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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)
Solution
Total time of journey = 6 s
g = 10 m/s2
Let 'H' be the greatest height.
Time of ascent, t = 6/2 = 3 s,
For ascent, initial velocity, u = 0
Using the second equation of motion,
H = ut + (1/2) gt2
H = 0 + (1/2) (10) (3) 2
H = 45 m
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