Question

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

• A. $\frac{g\left(t_{2}-t_{1}\right)^{2}}{8}$
• B. $\frac{g\left(t_{2}+t_{1}\right)^{2}}{4}$
• C. $\frac{g\left(t_{2}+t_{1}\right)^{2}}{8}$
• D. $\frac{g\left(t_{2}-t_{1}\right)^{2}}{4}$

Answer 1

Answer: (C)

When the body is projected vertically upwards with velocity u it occupies the same position while going up and coming down after time of $t_1$ and $t_2$.
$\therefore \quad H = ut -\frac{1}{2}gt^{2}$
$gt^{2} - 2ut + 2H = 0$
It is a quadratic equation in t, and $t_{1}$ and $t_{2}$ are the two roots of this equation.
$\therefore \quad$ Sum of roots $= t_{1} + t_{2} = − \left(\frac{-2u}{g}\right) = \frac{2u}{g}$
or $u = \frac{g\left(t_{1}-t_{2}\right)}{2}$
The maximum height attained is
$H_{max} = \frac{u^{2}}{2g} = \frac{1}{2g} \left[\frac{g\left(t_{1}+t_{2}\right)}{2}\right]^{2} = \frac{g\left(t_{1}+t_{2}\right)^{2}}{8}$