Projectile motion

Range of a projectile

R = u² sin 2θ / g. At 20 m/s and 15° with g = 10 m/s², the range is 20 m.

Projectile motion

Drag the launch arrow, or type the speed and angle. Slide the time to see the velocity at that moment. The dots are equal steps of time apart. Air resistance is ignored, as in NCERT.

Time 0.26 s

Time of flight T1.04 s2u sin θ / g

Range R20 mu² sin 2θ / g

Max height H1.34 mu² sin²θ / 2g

Velocity at 0.26 s19.49 m/svx = 19.32, vy = 2.59, 7.63° up

Pathy = 0.268x - 0.0134x²y = x tan θ - gx² / (2u² cos²θ)

A ball thrown at 20 m/s and 15 degrees rises 1.34 m, flies for 1.04 s and lands 20 m away.

Show the formulas

Time of flight: T = 2u sin θ / g = 2 × 20 × sin 15° / 10 = 1.04 s
Range: R = u² sin 2θ / g = 20² × sin 30° / 10 = 20 m
Max height: H = u² sin²θ / 2g = 20² × sin²15° / (2 × 10) = 1.34 m
Longest range: At θ = 45°, R max = u² / g = 40 m. θ and 90° - θ give the same range.
Velocity: vx = u cos θ = 19.32 m/s, vy = u sin θ - gt = 2.59 m/s at t = 0.26 s, so v = 19.49 m/s
Path: y = x tan θ - gx² / (2u² cos²θ), here y = 0.268x - 0.0134x²

Working

Step by step

Given u = 20 m/s at θ = 15°, with g = 10 m/s².
R = u² sin 2θ / g = 20² × sin 30° / 10 = 20 m
The ball is up for T = 2u sin θ / g = 1.04 s and moves sideways at u cos θ = 19.32 m/s, so it covers 19.32 × 1.04 ≈ 20 m.
At the same speed the longest range is at 45°: R max = u² / g = 20² / 10 = 40 m.

What the angle does

Range and height at 20 m/s

15° R = 20 mH = 1.34 m, T = 1.04 s
30° R = 34.64 mH = 5 m, T = 2 s
45° R = 40 mH = 10 m, T = 2.83 s
60° R = 34.64 mH = 15 m, T = 3.46 s
75° R = 20 mH = 18.66 m, T = 3.86 s

Derivation

Why it works

The ball flies for T = 2u sin θ / g, and the horizontal velocity u cos θ stays the same the whole time.
So the range is R = u cos θ × T = 2u² sin θ cos θ / g.
With 2 sin θ cos θ = sin 2θ this becomes R = u² sin 2θ / g.
sin 2θ is at most 1, at θ = 45°, so the longest range is u² / g.
Doubling the speed makes the range four times as long, because R grows with u².

Questions