From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the…

1. 2 g H = n²u²
2. g H = (n - 2)²u²
3. 2g H = nu²(n - 2)
4. g H = (n - 2)u²

Solution

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

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

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

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

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

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

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

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

2g H = nu²(n - 2)

Asked in: JEE Main 2014 (06 Apr)

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