Question 24
Two stones are thrown vertically upwards simultaneously with their initial velocities $u_{1}$ and $u_{2}$ respectively. Prove that the heights reached by them would be in the ration of $u_{1}^{2}$ ; $u_{2}^{2}$ ( Assume upward acceleration is –g and downward acceleration to be +g).

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Solution

We know for upward motion, $v^{2} = u^{2} - 2 g h$ or $h = \frac{u^{2} - v^{2}}{2 g}$
But at highest point $v = 0$
Therefore, $h = \frac{u^{2}}{2 g}$
For first ball, $h_{2} = \frac{u_{1}^{2}}{2 g}$
and for second ball $h_{2} = \frac{u_{2}^{2}}{2 g}$
Thus $\frac{h_{1}}{h_{2}} = \frac{\frac{u_{1}^{2}}{2 g}}{\frac{u_{2}^{2}}{2 g}} = \frac{u_{1}^{2}}{u_{1}^{2}} o r h_{1} : h_{2} = u_{1}^{2} : u_{2}^{2}$

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Archimedes' Principle

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Question 24 Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ration of u21 ; u22 ( Assume upward acceleration is –g and downward acceleration to be +g).

We know for upward motion, v2=u2−2gh or h=u2−v22g But at highest point v=0 Therefore, h=u22g For first ball, h2=u212g and for second ball h2=u222g Thus h1h2=u212gu222g=u21u21 or h1:h2=u21:u22

Question 24
Two stones are thrown vertically upwards simultaneously with their initial velocities $u_{1}$ and $u_{2}$ respectively. Prove that the heights reached by them would be in the ration of $u_{1}^{2}$ ; $u_{2}^{2}$ ( Assume upward acceleration is –g and downward acceleration to be +g).