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 $t_2$.
$\therefore H=ut-\frac{1}{2}g{t}^2$
$g{t}^2-2ut+2H=0$
It is a quadratic equation in t, and $t_1$ and $t_2$ are the two roots of this equation.
$\therefore$ 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
$H_{max}=\frac{u^2}{2g}=\frac{1}{2g}{\left[\frac{g\left(t_1+t_2\right)}{2}\right]}^2=\frac{g{\left(t_1+t_2\right)}^2}{8}$