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JIPMER 2017
Exam
JIPMER
Year
2017
Subject
Physics
Chapter
Motion in One Dimension

JIPMER 2017 · Physics · Motion in One Dimension

A body is projected vertically upwards. The times corresponding to...

A body is projected vertically upwards. The times corresponding to height hh while ascending and while descending are t1t_1 and t2t_2, respectively. Then, the velocity of projection will be (take gg as acceleration due to gravity)

Options

  1. Agt1t22\frac{g \sqrt{t_1 t_2}}{2}
  2. B\frac{g\left(t_1+t_2\right)}{2}
  3. Cgt1t2g \sqrt{t_1 t_2}
  4. Dg \frac{t_1 t_2}{\left(t_1+t_2\right)}

Correct answer

B. \frac{g\left(t_1+t_2\right)}{2}

Step-by-step solution

Let vv be initial velocity of vertical projection and tt be the time taken by the body to reach a height hh from ground. Here, u=u,a=−g,s=h,t=tu=u, a=-g, s=h, t=t Using, s=ut+12at2s=u t+\frac{1}{2} a t^2, we have h=ut+12(−g)t2 or  gt2−2ut+2h=0&∴t=2u±4u2−4g×2h2g=u±u2−2ghg\begin{aligned}& h=u t+\frac{1}{2}(-g) t^2 \text{\text{ or } } g t^2-2 u t+2 h=0 \& \therefore 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 tt has two values, i.e. t1=u+u2−2ghgt2=u−u2−2ghg&t1+t2=2ug or  u=g(t1+t2)2\begin{aligned}& t_1=\frac{u+\sqrt{u^2-2 g h}}{g} \\ & t_2=\frac{u-\sqrt{u^2-2 g h}}{g} \& t_1+t_2=\frac{2 u}{g} \text{\text{ or } } u=\frac{g\left(t_1+t_2\right)}{2}\end{aligned}
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