Question

A projectile has an initial velocity $v_0$ at an angle $\theta$ above the horizontal. It reaches the highest point of its trajectory in time $T$ after the launch. The highest point is at a vertical distance $H$ and at the horizontal distance $d$ from the point of projection. The speed of projectile at its highest point is $v$. For this situation, mark out the correct statement(s).

• A. $d=v_0\cos \theta \times T$
• B. $v=v_0\cos \theta$
• C. $H=\frac{{\left(v_0\sin \theta \right)}^2}{2g}$
• D. $H=\frac{g{T}^2}{2}$

Answer 1

Answer: (A), (B), (C), (D)

$H=\frac{g{T}^2}{2}$

Explanation for correct answer:

The velocity of the projectile along the horizontal direction is,

$v_x=v_{\circ }\cos \left(\theta \right)$

The velocity of the projectile along a vertical direction is,

$v_y=v_{\circ }\sin \left(\theta \right)$

Option A:

The equation of motion for the horizontal direction is $d=v_0\cos \theta \times T$

Option B:

The equation for the horizontal component of velocity at time $T$ is $v=v_0\cos \theta$

Option C:

The equation for the maximum height of the projectile is $H=\frac{{\left(v_0\sin \theta \right)}^2}{2g}$

Option D:

The equation for vertical motion, $H=\frac{g{T}^2}{2}$

Hence, option A, option B, option C, option D are correct.