A body is projected vertically upwards with a velocity $u$ from the top of a tower. Time taker by it to…

1. $\frac{n u^2(n-1)}{2 g}$
2. $\frac{n u^2(n-2)}{g}$
3. $\frac{n u^2(n-2)}{2 g}$
4. $\frac{u^2}{2 g}(n+1)$

Solution

Let the time taken to reach the maximum height, when thrown vertically upwards $ t_1=\frac{u}{g} $ If $t_2$ be the time to hit the ground, then given $ t_2=n t_1=n u / g $
$ \begin{aligned} & =\log \left|\frac{\cos 2 x}{1+\cos 2 x}\right| \\ \text { Now, IF } & =e^{\log \left|\frac{\cos 2 x}{1+\cos 2 x}\right|} \\ & =\frac{\cos 2 x}{1+\cos 2 x}=\frac{\cos 2 x}{2 \cos ^2 x} \end{aligned} $ Now, solution of differential equation is $ \begin{aligned} & \text { y. } \mathrm{IF}=\int(\theta . \mathrm{IF}) d x \\ & \Rightarrow y \frac{\cos 2 x}{2 \cos ^2 x}=\int \cos ^2 x \cdot \frac{\cos 2 x}{2 \cos ^2 x} d x \\ & \Rightarrow y\left(\frac{\cos 2 x}{\cos ^2 x}\right)=\int \cos 2 x d x \\ & \Rightarrow y\left(\frac{1-\tan ^2 x}{1+\tan ^2 x}\right) \frac{1}{\cos ^2 x}=\frac{\sin 2 x}{2}+c_1 \\ & \Rightarrow \quad y \frac{\left(1-\tan ^2 x\right)}{\sec ^2 x \cdot \cos ^2 x}=\frac{\sin 2 x}{2}+c_1 \\ & \Rightarrow \quad y\left(1-\tan ^2 x\right)=\frac{\sin 2 x+2 c_1}{2} \\ & \Rightarrow \quad y=\frac{\sin 2 x+c}{2\left(1-\tan ^2 x\right)} \\ & \end{aligned} $ So, using equation of motion for distance, i.e. $ \begin{aligned} s & =u t+\frac{1}{2} a t^2 \\ -H & =u \cdot n\left(\frac{u}{g}\right)-\frac{1}{2} g \frac{n^2 u^2}{g^2} \end{aligned} $ $ \text { [ } \left.s=-H, t=t_2=n t_1\right] $ or $ -H=\frac{n u^2}{g}-\frac{n^2 u^2}{2 g} $ or $ \begin{aligned} 2 g H & =n u^2(n-2) \\ H & =\frac{n u^2(n-2)}{2 g} \end{aligned} $

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

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