Q119 / 155 · WBJEE 2026 · Physics
From a tower of height $H$, a particle is thrown vertically upwards with a speed $u$. The time taken by the particle to hit the ground is $n$ times that taken by it to reach the highest point of its path. The relation between $H, u$ and $n$ is
- A$2gH = n^2u^2$
- B$gH = (n - 2)^2u^3$
- Correct answerC$2gH = nu^2(n - 2)$
- D$2gH = u^2(n - 2)^2$
Solution
Time taken by the particle to reach the highest point is given by $v = u - gt$.
At the highest point, $v = 0$, so $t_1 = \dfrac{u}{g}$.
The time taken to hit the ground is $t_2 = n t_1 = \dfrac{nu}{g}$.
Using the equation of motion $S = ut - \dfrac{1}{2}gt^2$ for the complete journey, taking upward direction as positive:
$-H = u t_2 - \dfrac{1}{2}g t_2^2$
Substituting $t_2 = \dfrac{nu}{g}$:
$-H = u \left( \dfrac{nu}{g} \right) - \dfrac{1}{2}g \left( \dfrac{nu}{g} \right)^2$
$-H = \dfrac{nu^2}{g} - \dfrac{n^2u^2}{2g}$
Multiplying the entire equation by $-2g$:
$2gH = n^2u^2 - 2nu^2$
$2gH = nu^2(n - 2)$
Answer: $2gH = nu^2(n - 2)$
At the highest point, $v = 0$, so $t_1 = \dfrac{u}{g}$.
The time taken to hit the ground is $t_2 = n t_1 = \dfrac{nu}{g}$.
Using the equation of motion $S = ut - \dfrac{1}{2}gt^2$ for the complete journey, taking upward direction as positive:
$-H = u t_2 - \dfrac{1}{2}g t_2^2$
Substituting $t_2 = \dfrac{nu}{g}$:
$-H = u \left( \dfrac{nu}{g} \right) - \dfrac{1}{2}g \left( \dfrac{nu}{g} \right)^2$
$-H = \dfrac{nu^2}{g} - \dfrac{n^2u^2}{2g}$
Multiplying the entire equation by $-2g$:
$2gH = n^2u^2 - 2nu^2$
$2gH = nu^2(n - 2)$
Answer: $2gH = nu^2(n - 2)$
Hence, the correct answer is option: 'C'.