Physics

The Geometry of Projectile Motion

We break down the parabolic path of a projectile into two independent components—horizontal and vertical—to find the range of a ball launched at 30 degrees.

u_x = u\cos\theta, \quad u_y = u\sin\theta

Think of the ball's motion not as one curve, but as two separate stories happening at once. The horizontal motion remains constant, while gravity pulls the vertical motion into a graceful arc.

Step-by-step solver

(a) Resolve initial velocity

Calculate horizontal and vertical components using the given launch speed and angle.

u_x = 20\cos 30^\circ = 10\sqrt{3}\ \text{m/s}, \ u_y = 20\sin 30^\circ = 10\ \text{m/s}

(b) Find time of flight

Set the vertical displacement to zero and solve for the time taken until the ball hits the ground.

0 = 10t - 4.9t^2 \implies t = \frac{20}{9.8} = \frac{100}{49}\ \text{s}

Multiply the constant horizontal velocity by the time of flight.

R = u_x \cdot t = 10\sqrt{3} \cdot \frac{100}{49} = \frac{1000\sqrt{3}}{49}\ \text{m} \approx 35.35\ \text{m}

Original question

A ball is thrown from ground level at 20 m/s at an angle of 30 degrees above the horizontal. Find the horizontal range of the ball. Take g = 9.8 m/s^2.

Follow-up chat

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