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  • From a Building of height H, a particle is projected vertically upwards with a speed uThe time taken by the particle, to h

From a Building of height H, a particle is projected vertically upwards with a speed uThe 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 of H, u, n is

Option: 1

gH\ =\left ( n-2 \right )u^{2}


Option: 2

2gH\ =\ n\ 2u^{2}


Option: 3

gH\ =\left ( n-2 \right )2u^{2}


Option: 4

2gH\ =\ nu^{2}(n-2)


Answers (1)

best_answer

 

Time taken to reach the maximum height

\begin{aligned} & \mathrm{V}=\mathrm{u}+\mathrm{at}_1 \\ & \Rightarrow \quad 0=\mathrm{u}-\mathrm{gt}_1 \\ & \Rightarrow \quad t_1=\frac{\mathrm{u}}{\mathrm{g}} \\ & \end{aligned}

Time taken to hit the ground

\begin{aligned} & \mathrm{s}=\mathrm{ut}_2+\frac{1}{2} \mathrm{at}_2^2 \quad[\mathrm{~s}=-\mathrm{H}] \\ & -\mathrm{H}=\mathrm{ut}_2-\frac{1}{2} \mathrm{gt}_2^2 \end{aligned}

\text { Given that } \mathrm{t}_2=\mathrm{nt}_1=\frac{\mathrm{nu}}{\mathrm{g}}

\begin{aligned} & -\mathrm{H}=\mathrm{u} \times \frac{\mathrm{nu}}{\mathrm{g}}-\frac{1}{2} \mathrm{~g} \times \frac{\mathrm{n}^2 \mathrm{u}^2}{\mathrm{~g}^2} \\ & \Rightarrow \quad 2 \mathrm{gH}=\mathrm{nu}^2(\mathrm{n}-2) \end{aligned}

 

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