A stone is dropped from the top of a tower and it strike with $3 k m / h r$ , against the ground. Another stone is thrown vertically downwards from the same top of the tower with a velocity $4 k m / h r$ . Its velocity when it strikes the ground will be:

7.0 k m / h r

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5.0 k m / h r

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3.5 k m / h r

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4.0 k m / h r

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Solution

The correct option is B $5.0 k m / h r$
$T h e v e l o c i t y o f f i r s t s t o n e o n r e a c h i n g g r o u n d i n t h e m / s i s 3 \times \frac{5}{18} = \frac{5}{6} m / s S o v^{2} = 2 g h \Rightarrow {(\frac{5}{6})}^{2} = 2 g h w h e r e h i s t h e h e i g h t o f t h e t o w e r, N o w f o r t h e s e c o n d s t o n e t h e u p w a r d i n i t i a l s p e e d = d o w n w a r d i n i t i a l s p e e d = 4 \times \frac{5}{18} = \frac{10}{9} m / s v^{2} = u^{2} + 2 g h \Rightarrow v^{2} = {(\frac{5}{6})}^{2} + {(\frac{10}{9})}^{2} = 1.929 {(m / s)}^{2} \times {(\frac{18}{5})}^{2} = 25 \Rightarrow v = 5 k m / h$

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