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Q.

A ball is thrown vertically upward with speed 10 m/s and return to ground with speed 8 m/s. Find the maximum height attained by ball. Assume the air resistance same throughout motion $(g = 10 {ms}^{- 2})$

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a

3.1 m

b

4.1 m

c

6.1 m

d

2.1 m

answer is C.

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Detailed Solution

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Upward motion : $K_{1} + U_{1} = K_{2} + U_{2} + W_{R}$
$\frac{1}{2} {m (10)}^{2} = mgh + Rh$
Downward motion :
$\begin{array}{l} K_{1} + U_{1} = K_{2} + U_{2} + W_{R} \\ mgh = \frac{1}{2} m (8)^{2} + Rh \end{array}$
Subtracting (ii) from (i), we get
$\begin{array}{l} \frac{1}{2} m (10)^{2} - mgh = mgh - \frac{1}{2} m (8)^{2} \\ 4 gh = (10)^{2} + (8)^{2} = 164 \\ h = \frac{164}{40} = 4.1 m \end{array}$

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