Projectile motion — find range, max height, and time of flight at angle θ

Question

A projectile is launched from the ground with initial velocity $u = 20\,\text{m/s}$ at an angle $\theta = 60°$ with the horizontal. Find the range, maximum height, and time of flight. (Take $g = 10\,\text{m/s}^2$)

(NCERT Class 11, Chapter 4)

Solution — Step by Step

Why This Works

Projectile motion is two independent motions: constant velocity along the horizontal and constant acceleration ($g$ downward) along the vertical. The time of flight is purely determined by the vertical motion — the projectile rises and falls. The range depends on both the horizontal speed and the total time in the air.

Maximum range occurs at $\theta = 45°$, where $\sin 2\theta = 1$. At $60°$, we get the same range as at $30°$ ($\sin 120° = \sin 60°$), but the $60°$ trajectory goes higher and stays in the air longer.

Alternative Method — Using kinematic equations directly

At maximum height, $v_y = 0$: $0 = u_y - gt_{\text{up}}$, so $t_{\text{up}} = u_y/g = 10\sqrt{3}/10 = \sqrt{3}\,\text{s}$.

$T = 2t_{\text{up}} = 2\sqrt{3}\,\text{s}$

$H = u_y t_{\text{up}} - \frac{1}{2}g t_{\text{up}}^2 = 10\sqrt{3} \times \sqrt{3} - \frac{1}{2} \times 10 \times 3 = 30 - 15 = 15\,\text{m}$

$R = u_x \times T = 10 \times 2\sqrt{3} = 20\sqrt{3}\,\text{m}$

Common Mistake