Two stones are thrown vertically upwards simultaneously with their initial velocities $u_{1}$ and $u_{2}$ respectively. The heights reached by them would be in the ratio of

u_{1}^{2} : u_{2}^{2}

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u_{1} : u_{2}

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u_{2}^{2} : u_{2}^{2}

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u_{2} : u_{1}

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Solution

The correct option is C $u_{1}^{2} : u_{2}^{2}$
$v^{2} = u^{2} + 2 a s$
$0 = (u_{1})^{2} + 2 (- g) s_{1}$
Hence, $s_{1} = u_{1}^{2} / 2 g$
Similarly, $s_{2} = u_{2}^{2} / 2 g$

Therefore, $\frac{s_{1}}{s_{2}} = \frac{u_{1}^{2}}{u_{2}^{2}}$

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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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Q. Show that

\frac{1}{u_{0}} - \frac{1}{u_{1}} + \frac{1}{u_{2}} - \frac{1}{u_{3}} + \dots + {(- 1)}^{n} \frac{1}{u_{n}} = \frac{1}{u_{0} +} \frac{u_{0}^{2}}{u_{1} - u_{0} +} \frac{u_{1}^{2}}{u_{2} - u_{1} +} \dots \frac{u_{n - 1}^{2}}{u_{n} - u_{n - 1}}

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Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. The heights reached by them would be in the ratio of

The correct option is C u21:u22v2=u2+2as0=(u1)2+2(−g)s1Hence, s1=u21/2gSimilarly, s2=u22/2gTherefore, s1s2=u21u22

Two stones are thrown vertically upwards simultaneously with their initial velocities $u_{1}$ and $u_{2}$ respectively. The heights reached by them would be in the ratio of

The correct option is C $u_{1}^{2} : u_{2}^{2}$
$v^{2} = u^{2} + 2 a s$
$0 = (u_{1})^{2} + 2 (- g) s_{1}$
Hence, $s_{1} = u_{1}^{2} / 2 g$
Similarly, $s_{2} = u_{2}^{2} / 2 g$

Therefore, $\frac{s_{1}}{s_{2}} = \frac{u_{1}^{2}}{u_{2}^{2}}$