A ball thrown up vertically returns to the thrower after 8s, its height above the ground after 6s. will be

20 m

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40 m

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60 m

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80 m

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Solution

The correct option is C 60 m
Time of Flight $= \frac{2 u}{g} = 8 s$
$\Rightarrow u = 40 m / s$
Its height after 6s $= u t - \frac{1}{2} g t^{2}$
$= 40 (6) - 5 (6)^{2} = 60 m$
Therefore, C is correct option.

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Similar questions

A ball is thrown up vertically returns to the thrower after 8s. Its heights above the ground after 6s will be ( take g= 10m/s^2)

A ball thrown up vertically returns to the thrower after 6 s. Find

(a) the velocity with which it was thrown up,

(b) the maximum height it reaches, and

6 s

4 s

9.8 m s^{- 2}

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