A ball is thrown at 30° with speed 20 m/s — find range and max height

Question

A ball is thrown with a speed of 20 m/s at an angle of 30° with the horizontal. Find (a) the maximum height reached and (b) the horizontal range. (Take $g = 10\ \text{m/s}^2$ )

Solution — Step by Step

The initial velocity $u = 20\ \text{m/s}$ at angle $\theta = 30°$ .

Horizontal component: $u_x = u\cos 30° = 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3}\ \text{m/s}$

Vertical component: $u_y = u\sin 30° = 20 \times \dfrac{1}{2} = 10\ \text{m/s}$

These two components act independently — horizontal is uniform motion, vertical is uniformly decelerated by gravity.

At maximum height, the vertical velocity becomes zero. Using $v^2 = u_y^2 - 2gH$ :

0 = (10)^2 - 2 \times 10 \times H

H = \frac{100}{20} = 5\ \text{m}

Alternatively, use the direct formula:

H = \frac{u^2 \sin^2\theta}{2g} = \frac{400 \times \sin^2 30°}{20} = \frac{400 \times 0.25}{20} = \frac{100}{20} = 5\ \text{m}

Maximum height = 5 m

We need the total time the ball is in the air to find the range.

T = \frac{2u_y}{g} = \frac{2 \times 10}{10} = 2\ \text{s}

By symmetry, the ball takes equal time going up and coming down.

Since horizontal velocity is constant (no air resistance), range = horizontal velocity × time of flight:

R = u_x \times T = 10\sqrt{3} \times 2 = 20\sqrt{3}\ \text{m} \approx 34.6\ \text{m}

Or use the direct formula:

R = \frac{u^2 \sin 2\theta}{g} = \frac{400 \times \sin 60°}{10} = \frac{400 \times \frac{\sqrt{3}}{2}}{10} = \frac{200\sqrt{3}}{10} = 20\sqrt{3}\ \text{m}

(a) Maximum height = 5 m

(b) Horizontal range = $20\sqrt{3} \approx 34.6\ \text{m}$

Why This Works

The key insight is that projectile motion is just two independent one-dimensional motions happening simultaneously. The horizontal direction has no force (constant velocity), while the vertical direction is under constant gravitational acceleration downward.

Because of this independence, we can apply standard kinematics equations to each direction separately — then combine the results. The projectile traces a parabolic path precisely because horizontal distance grows uniformly while vertical distance grows as $t^2$ .

Alternative Method

You can also find the time to reach maximum height directly: $t_{up} = \dfrac{u_y}{g} = \dfrac{10}{10} = 1\ \text{s}$ , then total time = $2 \times 1 = 2\ \text{s}$ . Range = $u_x \times 2t_{up}$ . Same result.

Remember $\sin 2\theta = 2 \sin\theta \cos\theta$ — the range formula $R = \dfrac{u^2 \sin 2\theta}{g}$ is maximum when $\theta = 45°$ . For 30° and 60°, the range is the same! This is a favourite CBSE conceptual question.

Common Mistake

Students often forget to use $\sin 2\theta$ in the range formula and instead write $\sin\theta$ . Note that $\sin 2(30°) = \sin 60° = \dfrac{\sqrt{3}}{2}$ , not $\sin 30° = \dfrac{1}{2}$ . Confusing these halves your calculated range.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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