Time taken by an object falling from rest to cover the height of $h_1$ and $h_2$ is respectively $t_1$ and $t_

Question

Time taken by an object falling from rest to cover the height of $h_1$ and $h_2$ is respectively $t_1$ and $t_2$. Then the ratio of $t_1$ to $t_2$ is:

Accepted Answer

1. **Equation of Motion:** For an object falling freely from rest ($u=0$) under gravity ($a=g$), the distance covered $h$ in time $t$ is given by the second equation of motion:
   $$ h = ut + \frac{1}{2}gt^2 $$
   $$ h = 0 + \frac{1}{2}gt^2 \implies h = \frac{1}{2}gt^2 $$
2. **Relation between Time and Height:** Rearranging the equation for time:
   $$ t = \sqrt{rac{2h}{g}} $$
   Since $g$ is constant, $t \propto \sqrt{h}$.
3. **Calculate Ratio:**
   - For height $h_1$, time $t_1 = \sqrt{rac{2h_1}{g}}$
   - For height $h_2$, time $t_2 = \sqrt{rac{2h_2}{g}}$
   - Ratio $\frac{t_1}{t_2} = \frac{\sqrt{2h_1/g}}{\sqrt{2h_2/g}} = \sqrt{\frac{h_1}{h_2}}$ .

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