Two bodies of different masses $m_a$ and $m_b$ are dropped from two different heights $a$ and $b$. The ratio o

Question

Two bodies of different masses $m_a$ and $m_b$ are dropped from two different heights $a$ and $b$. The ratio of the time taken by the two to cover these distances is:

Accepted Answer

1. **Identify the Physics Principle:** For a body dropped from rest ($u=0$) under gravity, the motion is uniformly accelerated with acceleration $g$. The time of fall is independent of the mass of the body (Galileo's observation) .
2. **Apply Kinematic Equation:** Use the second equation of motion, $s = ut + \frac{1}{2}at^2$. 
   - For the first body: Height $= a$, Time $= t_1$.
     $$ a = 0 + \frac{1}{2}gt_1^2 \implies t_1 = \sqrt{\frac{2a}{g}} $$
   - For the second body: Height $= b$, Time $= t_2$.
     $$ b = 0 + \frac{1}{2}gt_2^2 \implies t_2 = \sqrt{\frac{2b}{g}} $$
3. **Calculate Ratio:**
   $$ \frac{t_1}{t_2} = \frac{\sqrt{2a/g}}{\sqrt{2b/g}} = \frac{\sqrt{a}}{\sqrt{b}} $$
   Thus, the ratio is $\sqrt{a} : \sqrt{b}$ .

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