A body is projected vertically upwards at time $t=0$ and it is seen at a height $H$ at time $t_1$ and $t_2$ second during its flight. The maximum height attained is ( $g$ is acceleration due to gravity)

Question

Aicharya · Accepted Answer

Let time taken by the body to fal from point $C$ to $B$ is $t^{\prime}$. Then $t_1+2 t^{\prime}=t_2$

$t^{\prime}=\left(\frac{t_2-t_1}{2}\right) \ldots(\mathrm{i})$ Total time taken to reach point $C$ $\begin{aligned} T & =t_1+t^{\prime} \\ & =t_1+\frac{t_2-t_1}{2} \\ & =\frac{2 t_1+t_2-t_1}{2} \\ & =\left(\frac{t_1+t_2}{2}\right) \end{aligned}$ Maximum height attained $\begin{aligned} H_{\max } & =\frac{1}{2} g(T)^2 \\ & =\frac{1}{2} g\left(\frac{t_1+t_2}{2}\right)^2 \\ & =\frac{1}{2} g \cdot \frac{\left(t_1+t_2\right)^2}{4} \\ \Rightarrow \quad H_{\max } & =\frac{1}{8} g \cdot\left(t_1+t_2\right)^2 \mathrm{~m} \end{aligned}$

A body is projected vertically upwards at time $t=0$ and it is seen at a height $H$ at time $t_1$ and $t_2$…

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