C R UI 20 ms. How far below the top...
Question  # 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

JEE/Engineering Exams
Physics
Solution 138 4.0 (1 ratings)  ( 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} ] Quick and Stepwise Solutions Just click and Send OVER 20 LAKH QUESTIONS ANSWERED Download App for Free