Q159 / 180 · COMEDK 2023 (Memory Based) · Physics

A body is projected vertically upwards. The times corresponding to height $h$ while ascending and while descending are $t_1$ and $t_2$, respectively. Then, the velocity of projection will be (take, $g$ as acceleration due to gravity)

  • A
    $\frac{g \sqrt{t_1 t_2}}{2}$
  • Correct answer
    B
    $\frac{g\left(t_1+t_2\right)}{2}$
  • C
    $g \sqrt{t_1 t_2}$
  • D
    $g \frac{t_1 t_2}{\left(t_1+t_2\right)}$

Solution

Let $v$ be initial velocity of vertical projection and $t$ be the time taken by the body to reach a height $h$ from ground. Here, $\quad u=u, a=-g, s=h, t=t$ Using, $s=u t+\frac{1}{2} a t^2$, we have $\begin{aligned} & h=u t+\frac{1}{2}(-g) t^2 \\ & g t^2-2 u t+2 h=0 \\ & t=\frac{2 u \pm \sqrt{4 u^2-4 g \times 2 h}}{2 g}=\frac{u \pm \sqrt{u^2-2 g h}}{g} \end{aligned}$ It means $t$ has two values, i.e. $\begin{aligned} t_1 & =\frac{u+\sqrt{u^2-2 g h}}{g} \Rightarrow t_2=\frac{u-\sqrt{u^2-2 g h}}{g} \\ t_1+t_2 & =\frac{2 u}{g} \text { or } u=\frac{g\left(t_1+t_2\right)}{2} \end{aligned}$

Hence, the correct answer is option: 'B'.

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