Two stones are thrown vertically upwards simultaneously with their initial velocities $u_{1}$ and $u_{2}$ respectively. What is ratio of the heights reached by them (Assume acceleration is g downwards)?

u_{1} : u_{2}

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

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u_{1}^{3} : u_{2}^{3}

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None of the above

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Solution

The correct option is B $u_{1}^{2} : u_{2}^{2}$
We know for upward motion $v^{2} = u^{2} - 2 g h$
$\Rightarrow h = \frac{u^{2} - v^{2}}{2 g}$
$At highest point, v = 0$
$∴ h_{m a x} = \frac{u^{2}}{2 g}$
For first ball, $h_{1} = \frac{u_{1}^{2}}{2 g}$
For second ball, $h_{2} = \frac{u_{2}^{2}}{2 g}$
Thus, $\frac{h_{1}}{h_{2}} = \frac{u_{1}^{2}}{u_{2}^{2}}$
$\Rightarrow h_{1} : h_{2} = u_{1}^{2} : u_{2}^{2}$

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Two stones are thrown up simultaneously with initial speeds of $u_{1}$ and $u_{2} (u_{2} > u_{1})$ . They hit the ground after 6 s and 10 s respectively. Which of the following graphs correctly represents the time variation of $Δ x = (x_{2} - x_{1})$ , the relative position of the second stone with respect to the first till t = 10 s? Assume that the stones do not rebound after hitting the ground.