$A$ particle is projected with velocity $2 \sqrt{gh}$ at an angle $60^{\circ}$ to the horizontal so that it just clears two walls of equal height $h$ which are at a distance $2h$ from each other. The time taken by the particle to travel between these two walls is

  • A
    $2 \sqrt{\frac{2h}{g}}$
  • B
    $\sqrt{\frac{h}{2g}}$
  • C
    $2 \sqrt{\frac{h}{g}}$
  • D
    $\sqrt{\frac{h}{g}}$

Explore More

Similar Questions

$A$ projectile is projected with speed $u$ at an angle $\theta$ with the horizontal. The average velocity of the projectile between the instants it crosses the same level is ............

$A$ projectile is projected at $10 \ m/s$ by making an angle of $60^{\circ}$ to the horizontal. After some time,its velocity makes an angle of $30^{\circ}$ to the horizontal. Its speed at this instant is:

$A$ body is projected with a speed $u$ at an angle $\theta$ with the horizontal. The radius of curvature of the trajectory,when it makes an angle $\left(\frac{\theta}{2}\right)$ with the horizontal is ($g$ = acceleration due to gravity).

The ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is $x$. The angle of projection with the horizontal is

$A$ ball is thrown at different angles with the same speed $u$ and from the same point. It has the same range in both cases. If $y_1$ and $y_2$ are the heights attained in the two cases,then $y_1 + y_2$ is equal to:

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