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Two objects of mass $m_{1}$ and $m_{2}$ , respectively, are dropped simultaneously from heights $h_{1}$ and $h_{2}$ from the ground. The ratio of time taken by stone $1$ to reach the ground to the time taken by stone $2$ to reach the ground is

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\sqrt{h_{2}}

\sqrt{h_{1}}

\sqrt{h_{1}}

\sqrt{h_{2}}

answer is D.

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Detailed Solution

In the above case, the ratio of time taken by the stones is

\sqrt{h_{1}}

\sqrt{h_{2}}

.
As the stones are dropped, the initial velocity of both stones is zero. The acceleration of the stones is the acceleration due to gravity(g) in the downward direction.
Stone-1:
Here, Initial velocity

(u) = 0 m / s

Acceleration

(a) = g m / s^{2}

Distance travelled

(s) = h_{1}

Applying the equation of motion

s = ut + \frac{1}{2} a t^{2}

Putting the values

h_{1} = 0 + \frac{1}{2} \times g \times t_{1}^{2}

\Rightarrow t_{1} = \sqrt{\frac{2 h_{1}}{g}}

………(1) Stone-2: Here, Initial velocity

(u) = 0 m / s

Acceleration

(a) = g m / s^{2}

Distance travelled

(s) = h_{2}

Applying the equation of motion

s = ut + \frac{1}{2} a t^{2}

Putting the values

h_{2} = 0 + \frac{1}{2} \times g \times t_{2}^{2}

\Rightarrow t_{2} = \sqrt{\frac{2 h_{2}}{g}}

……..(2)
Diving equation (1) by (2)

\Rightarrow \frac{t_{1}}{t_{2}} = \frac{\sqrt{\frac{2 h_{1}}{g}}}{\sqrt{\frac{2 h_{2}}{g}}}

\Rightarrow

\frac{t_{1}}{t_{2}} = \frac{\sqrt{h_{1}}}{\sqrt{h_{2}}}

Hence, the ratio of time taken by stone 1 to reach the ground to the time taken by stone 2 to reach the ground is

\sqrt{h_{1}}

\sqrt{h_{2}}

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