Explanation

A projectile is launched at speed u and angle θ from heighth. Split the velocity and use constant acceleration — horizontala = 0, vertical a = −g — so range R, greatest height H, and time of flight T follow at once.h = 0 is level ground.

After launch the particle is free, so the two directions are independent. There is no string and no tension.

• Horizontal: x = (u cos θ) t, v_x = u cos θ
• Vertical: y = h + (u sin θ) t − ½g t², v_y = u sin θ − g t

Landing is y = 0. The physical root of that quadratic isT = (u sin θ + √((u sin θ)² + 2g h)) / g. Then R = (u cos θ) T andH = h + (u sin θ)² / (2g).

On level ground this collapses to T = 2u sin θ / gand R = u² sin 2θ / g, which is greatest at 45°. A horizontal throw is θ = 0 with h > 0.

Worked example

A particle is projected from ground level with speed 20 m/s at 45° to the horizontal. Take g = 9.81 m/s². Find the time of flight, the range, and the greatest height.

1. u cos 45° = u sin 45° = 14.14 m/s
2. Time of flight T = 2 × 14.14 / 9.81 = 2.88 s
3. Range R = 14.14 × 2.88 = 40.77 m
4. Greatest height H = 14.14² / (2 × 9.81) = 10.19 m

The same particle is projected from ground level at 30° to the horizontal, still at 20 m/s. Find T, R, and H.

1. u cos 30° = 17.32 m/s, u sin 30° = 10 m/s
2. T = 2 × 10 / 9.81 = 2.04 s
3. R = 17.32 × 2.04 = 35.31 m, and H = 100 / 19.62 = 5.10 m

On level ground the range is greatest at 45°. From a height, bothT and R increase.