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

Engineering

Physics

Straight Line Motion

Question

g H = (n – 2)u²

g H = (n – 2)²u²

2g H = nu²(n – 2)

2 g H = n²u²

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$- H = ut - \frac{- 1}{2} {gt}^{2}$

$\frac{1}{2}$ gt² – ut – H = 0

$\frac{nu}{g} = t = \frac{ut \sqrt{u^{2} + 2 gM}}{g}$

(n – 1)u = $\sqrt{u^{2} + 2 gH}$

(n – 1)² u² – u² = 2gH

u²[n² – 2n] = 2gH