Given that, the height achieved in 1s,h=25m Let u be the initial velocity in upward direction. By equation of motion, h=ut−
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gt2
25=u(1)−
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(10)(1)2 25=u−5⇒u=30m∕s Final velocity of ball at maximum height become instantly zero. ∴ Time taken to reach at maximum height is calculated as v=u−gt⇒0=u−10t⇒t=
30
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=3s Now, distance travelled in 2s= Displacement in 2s By equation of motion, d1=s1=ut−
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gt2 =30(2)−
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×10×(2)2 =60−5×4=60−20=40m Maximum height achieved by ball in (3s) H=ut−
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gt2=30(3)−
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×10(3)2 =90−
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×9 =90−45=45m Displacement after 4s, s2=ut−
