Projectile motion

Maximum height of a projectile

H = u² sin²θ / 2g. At 40 m/s and 30° with g = 10 m/s², the ball rises 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 1 s

Time of flight T4 s2u sin θ / g

Range R138.56 mu² sin 2θ / g

Max height H20 mu² sin²θ / 2g

Velocity at 1 s36.06 m/svx = 34.64, vy = 10, 16.1° up

Pathy = 0.577x - 0.00417x²y = x tan θ - gx² / (2u² cos²θ)

A ball thrown at 40 m/s and 30 degrees rises 20 m, flies for 4 s and lands 138.56 m away.

Show the formulas

Time of flight: T = 2u sin θ / g = 2 × 40 × sin 30° / 10 = 4 s
Range: R = u² sin 2θ / g = 40² × sin 60° / 10 = 138.56 m
Max height: H = u² sin²θ / 2g = 40² × sin²30° / (2 × 10) = 20 m
Longest range: At θ = 45°, R max = u² / g = 160 m. θ and 90° - θ give the same range.
Velocity: vx = u cos θ = 34.64 m/s, vy = u sin θ - gt = 10 m/s at t = 1 s, so v = 36.06 m/s
Path: y = x tan θ - gx² / (2u² cos²θ), here y = 0.577x - 0.00417x²

Working

Step by step

Given u = 40 m/s at θ = 30°, with g = 10 m/s².
Vertical launch speed uy = u sin θ = 40 × sin 30° = 20 m/s.
At the top vy = 0, so 0 = uy² - 2gH and H = uy² / 2g = 20² / (2 × 10).
H = u² sin²θ / 2g = 40² × sin²30° / (2 × 10) = 20 m
It gets there after t = uy / g = 2 s, half the time of flight.

Derivation

Why it works

Only the vertical part of the motion decides the height. It starts at uy = u sin θ and slows down at g.
At the highest point the vertical velocity is zero, though the ball still moves sideways at u cos θ.
Use v² = u² - 2gs vertically: 0 = u² sin²θ - 2gH.
So H = u² sin²θ / 2g.
It takes t = u sin θ / g to get there, which is half the time of flight.

Questions