Question 179 of 308

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 :

A2 g H = n2u2Bg H = (n - 2)2u2C2g H = nu2(n - 2)Dg H = (n - 2)u2

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Answer: (C) 2g H = nu2(n - 2)

${\text{nt}}_{\text{AB}}={\text{t}}_{\text{AB}}+{\text{t}}_{\text{BC}}+{\text{t}}_{\text{CD}}$

$\left(\text{n}-1\right){\text{t}}_{\text{AB}}={\text{t}}_{\text{BC}}+{\text{t}}_{\text{CD}}....(1)$

$\left(\text{n}-1\right)\dfrac{\text{u}}{\text{g}}=\dfrac{\text{u}}{\text{g}}+{\text{t}}_{\text{CD}}$

$∴\left(\text{n}-2\right)\dfrac{\text{u}}{\text{g}}={\text{t}}_{\text{CD}}$ ...(2)

$\text{H}+\text{h}=\dfrac{1}{2}{\text{gt}}_{\text{BD}}^{2}$

$\text{H}+\dfrac{{\text{u}}^{2}}{2\text{g}}=\dfrac{1}{2}\text{g}{\left[\dfrac{\left(\text{n}-1\right)\text{u}}{\text{g}}\right]}^{2}$ ...from (1)

$∴\text{H}+\dfrac{{\text{u}}^{2}}{2\text{g}}=\dfrac{{\left(\text{n}-1\right)}^{2}{\text{u}}^{2}}{2\text{g}}$

$∴\text{H}=\dfrac{{\text{u}}^{2}}{2\text{g}}\left[{\left(\text{n}-1\right)}^{2}-1\right]$

2g H = nu2(n - 2)