Numford
Projectile motion

Projectile thrown at 30 degrees

Thrown at 28 m/s and 30° with g = 9.8 m/s², the ball rises H = 10 m, stays up T = 2.86 s (about 2.9 s) and lands R = 69.28 m (about 69 m) away.

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 of flight T2.86 s2u sin θ / g
Range R69.28 mu² sin 2θ / g
Max height H10 mu² sin²θ / 2g
Velocity at 0.71 s25.24 m/svx = 24.25, vy = 7, 16.1° up
Pathy = 0.577x - 0.00833x²y = x tan θ - gx² / (2u² cos²θ)

A ball thrown at 28 m/s and 30 degrees rises 10 m, flies for 2.86 s and lands 69.28 m away.

g (m/s²)
Show the formulas
Time of flight
T = 2u sin θ / g = 2 × 28 × sin 30° / 9.8 = 2.86 s
Range
R = u² sin 2θ / g = 28² × sin 60° / 9.8 = 69.28 m
Max height
H = u² sin²θ / 2g = 28² × sin²30° / (2 × 9.8) = 10 m
Longest range
At θ = 45°, R max = u² / g = 80 m. θ and 90° - θ give the same range.
Velocity
vx = u cos θ = 24.25 m/s, vy = u sin θ - gt = 7 m/s at t = 0.71 s, so v = 25.24 m/s
Path
y = x tan θ - gx² / (2u² cos²θ), here y = 0.577x - 0.00833x²
Working

Step by step

  1. Given u = 28 m/s at θ = 30°, with g = 9.8 m/s².
  2. H = u² sin²θ / 2g = 28² × sin²30° / (2 × 9.8) = 10 m
  3. T = 2u sin θ / g = 2 × 28 × sin 30° / 9.8 = 2.86 s (about 2.9 s)
  4. R = u² sin 2θ / g = 28² × sin 60° / 9.8 = 69.28 m (about 69 m)
  5. A throw at 60° lands at the same spot, and goes three times as high.
What the angle does

Range and height at 28 m/s

  • 15° R = 40 mH = 2.68 m, T = 1.48 s
  • 30° R = 69.28 mH = 10 m, T = 2.86 s
  • 45° R = 80 mH = 20 m, T = 4.04 s
  • 60° R = 69.28 mH = 30 m, T = 4.95 s
  • 75° R = 40 mH = 37.32 m, T = 5.52 s
Derivation

Why it works

  1. Split the launch velocity: ux = u cos θ along the ground and uy = u sin θ upwards. Gravity only acts downwards, so ux never changes.
  2. Vertically, y = u sin θ t - ½gt². The ball is back on the ground when y = 0, so T = 2u sin θ / g.
  3. Horizontally, x = u cos θ t. In the time T it goes R = u cos θ × 2u sin θ / g = u² sin 2θ / g.
  4. At the top the vertical velocity is zero. From v² = u² - 2gs with v = 0, H = u² sin²θ / 2g.
  5. Taking t = x / (u cos θ) out of the y equation gives the path y = x tan θ - gx² / (2u² cos²θ), a parabola.
Questions

Projectile at 30 degrees: quick answers

What is the maximum height of a ball thrown at 28 m/s at 30°?

With g = 9.8 m/s², H = 28² × sin²30° / (2 × 9.8) = 10 m.

How long is it in the air and how far does it go?

T = 2 × 28 × sin 30° / 9.8 = 2.86 s (about 2.9 s) and R = 28² × sin 60° / 9.8 = 69.28 m (about 69 m).

Which other angle gives the same range?

60°, because the two add to 90 degrees. It goes three times as high.

Try your own throw