A body at rest falls through a height 'h' with velocity 'V'. If it has to fall down further for its velocity to become three times, the distance travelled in that interval is
Show Hint
Since distance scales with the square of velocity ($H \propto v^2$), tripling the velocity ($1 \to 3$) requires multiplying the total fall distance from rest by $3^2 = 9$. Subtracting the initial distance $1h$ gives an additional distance of $9h - 1h = 8h$ instantly.
Step 1: The story.
A body is dropped from rest. After falling a height $h$ its speed is $V$. We want the extra height it must fall so its speed becomes $3V$.
Step 2: The useful equation.
For a body starting from rest under gravity, the equation of motion gives
\[ v^2 = 2gH \]
where $H$ is the total fall from the start. So $v^2$ grows in step with the total height.
Step 3: First stage.
To reach speed $V$ over height $h$:
\[ V^2 = 2gh \]
Step 4: Stage to reach 3V.
Let $h'$ be the total fall from the start to reach $3V$:
\[ (3V)^2 = 2gh' \;\Rightarrow\; 9V^2 = 2gh' \]
Step 5: Compare the two.
Divide the second by the first:
\[ \frac{9V^2}{V^2} = \frac{2gh'}{2gh} \;\Rightarrow\; h' = 9h \]
So the total fall to reach $3V$ is $9h$.
Step 6: Find the extra distance.
The question asks only for the extra fall after speed $V$:
\[ \Delta h = h' - h = 9h - h = 8h \]
This is option (1).
\[ \boxed{\Delta h = 8h} \]