Question

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)

• A. $\frac{g(t_2-t_1{)}^2}{8}$
• B. $\frac{g(t_1+t_2{)}^2}{4}$
• C. $\frac{g(t_1+t_2{)}^2}{8}$
• D. $\frac{g(t_2-t_1{)}^2}{4}$

Answer 1

The correct option is D $\frac{g(t_2-t_1{)}^2}{4}$

time taken by the body from going to top after seen from height H is equal to time taken by the body from top to height H.

so time will be $\frac{(t_2-t_1)}{2}$.

total time to reach from ground to top is $t_1+\frac{(t_2-t_1)}{2}$=.$\frac{(t_2+t_1)}{2}$. time of flight is known we can take this in reverse direction initially at top with zero velocity and acceleration g, then find displacement after time of flight $\frac{(t_2+t_1)}{2}$

so $h=ut+\frac{1}{2}a{t}^2$=$\frac{1}{2}g(\frac{(t_2+t_1)}{2}{)}^2$

so option D is best possible answer.