From a tower of height H, a particle is throw | vedclass

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(B) Let $t_1$ be the time taken to reach the highest point. At the highest point, the final velocity is $0$. Using $v = u + at$, we get $0 = u - gt_1$

Vedclass · Accepted Answer

$2gH=nu^2(n-2)$

Vedclass · Answer

(B) Let $t_1$ be the time taken to reach the highest point. At the highest point, the final velocity is $0$. Using $v = u + at$, we get $0 = u - gt_1$, so $t_1 = \frac{u}{g}$.Let $T$ be the total time taken to hit the ground. Using the equation of motion $s = ut + \frac{1}{2}at^2$ with displacement $s = -H$, initial velocity $u$, and acceleration $a = -g$, we have:$-H = uT - \frac{1}{2}gT^2$$\frac{1}{2}gT^2 - uT - H = 0$Given $T = nt_1 = n(\frac{u}{g}) = \frac{nu}{g}$.Substituting $T$ into the equation:$\frac{1}{2}g(\frac{nu}{g})^2 - u(\frac{nu}{g}) - H = 0$$\frac{n^2u^2}{2g} - \frac{nu^2}{g} - H = 0$Multiply by $2g$:$n^2u^2 - 2nu^2 - 2gH = 0$$n^2u^2 - 2nu^2 = 2gH$$nu^2(n - 2) = 2gH$.

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:

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