$A$ body is projected vertically upwards at time $t=0$ and it is seen at a height $H$ at times $t_1$ and $t_2$ seconds 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
    $\frac{g(t_2-t_1)^2}{4}$

Explore More

Similar Questions

The initial and final velocities of a body projected vertically from the ground are $20 \,ms^{-1}$ and $18 \,ms^{-1}$ respectively. The maximum height reached by the body is (Acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,m$)

$A$ body is thrown vertically upwards with a velocity $u$. Find the true statement from the following:

Consider the acceleration,velocity,and displacement of a tennis ball as it falls to the ground and bounces back. The directions of which of these change during the process?

$A$ stone is thrown vertically upward with an initial velocity $v_0$. The distance travelled in time $t = \frac{1.5 v_0}{g}$ is

Difficult
View Solution

Four marbles are dropped from the top of a tower one after the other with an interval of one second. The first one reaches the ground after $4$ seconds. When the first one reaches the ground,the distances between the first and second,the second and third,and the third and fourth marbles will be respectively:

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo