Question

( u_{2} )
( S=15 mathrm{m} )
( v_{2}=0 )
( v_{2}^{2}-u_{2}^{2}=-2 g S Rightarrow 0-u_{2}^{2}=-2 times 10 times 15 )
( Rightarrow U_{2}=10 sqrt{3} mathrm{m} / mathrm{s} )
Now'
[
begin{array}{l}
forall phi eq U_{2}=frac{V_{1}}{2} eq V_{1}=20 sqrt{3} mathrm{m} / lambda
Rightarrow text { So; } V_{1}^{2}-u^{2}=2 times g times 15: Rightarrow(400 times 3)-4^{2}=2 times 10 times 15
end{array}
]
( Rightarrow u^{2}=1200-300 )
[
frac{u^{2}=900}{u=30 m / s}
]

# C R UI 20 ms. How far below the top of the cliff will the second stone overtake the hits A boy throws a ball with speed u in a well of depth 14 m as shown. On bounce with bottom of well the speed of the ball gets halved. What should be the minimum value of u in m/s) such that the ball may be able to reach his hand again? It is given that his hands are at I m height from top of the well while throwing and catching. 14 m

Solution