The speed of a projectile at its maximum height is $\frac{\sqrt{3}}{2}$ times its initial speed. If the range of the projectile is $P$ times the maximum height attained by it,$P$ is equal to

  • A
    $\frac{4}{3}$
  • B
    $2\sqrt{3}$
  • C
    $4\sqrt{3}$
  • D
    $\frac{3}{4}$

Explore More

Similar Questions

The maximum horizontal range of a projectile is $400\, m$. The maximum height attained by it will be ......... $m$.

An object is projected with a velocity of $20 \ m/s$ making an angle of $45^{\circ}$ with the horizontal. The equation for the trajectory is $h = Ax - Bx^2$, where $h$ is the height, $x$ is the horizontal distance, and $A$ and $B$ are constants. The ratio $A:B$ is $(g = 10 \ m/s^2)$.

Two particles $A$ and $B$ are projected simultaneously from a fixed point on the ground. Particle $A$ is projected on a smooth horizontal surface with speed $v$,while particle $B$ is projected in air with speed $\frac{2v}{\sqrt{3}}$. If particle $B$ hits particle $A$,the angle of projection of $B$ with the vertical is

Difficult
View Solution

The path of a projectile in the absence of air drag is shown in the figure by a dotted line. If air resistance is not ignored,then which one of the paths shown in the figure is appropriate for the projectile?

$A$ projectile is thrown at an angle $\theta$ such that it is just able to cross a vertical wall of height $H$ at its highest point,as shown in the figure. The horizontal distance from the point of projection to the wall is $\sqrt{3} H$. The angle $\theta$ at which the projectile is thrown is given by:

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