Numford
Projectile motion

Projectile thrown at 60 degrees

Thrown at 28 m/s and 60° with g = 9.8 m/s², the ball rises H = 30 m, stays up T = 4.95 s and lands R = 69.28 m away, as far as at 30°.

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 T4.95 s2u sin θ / g
Range R69.28 mu² sin 2θ / g
Max height H30 mu² sin²θ / 2g
Velocity at 1.24 s18.52 m/svx = 14, vy = 12.12, 40.89° up
Pathy = 1.73x - 0.025x²y = x tan θ - gx² / (2u² cos²θ)

A ball thrown at 28 m/s and 60 degrees rises 30 m, flies for 4.95 s and lands 69.28 m away.

g (m/s²)
Show the formulas
Time of flight
T = 2u sin θ / g = 2 × 28 × sin 60° / 9.8 = 4.95 s
Range
R = u² sin 2θ / g = 28² × sin 120° / 9.8 = 69.28 m
Max height
H = u² sin²θ / 2g = 28² × sin²60° / (2 × 9.8) = 30 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 = 14 m/s, vy = u sin θ - gt = 12.12 m/s at t = 1.24 s, so v = 18.52 m/s
Path
y = x tan θ - gx² / (2u² cos²θ), here y = 1.73x - 0.025x²
Working

Step by step

  1. Given u = 28 m/s at θ = 60°, with g = 9.8 m/s².
  2. H = u² sin²θ / 2g = 28² × sin²60° / (2 × 9.8) = 30 m
  3. T = 2u sin θ / g = 2 × 28 × sin 60° / 9.8 = 4.95 s
  4. R = u² sin 2θ / g = 28² × sin 120° / 9.8 = 69.28 m (about 69 m)
  5. A throw at 30° lands at the same spot, and goes a third 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. The range is R = u² sin 2θ / g.
  2. For the angle 90° - θ, sin 2(90° - θ) = sin(180° - 2θ) = sin 2θ.
  3. So θ and 90° - θ give exactly the same range, for example 15° and 75° or 30° and 60°.
  4. The steeper throw goes higher (H grows with sin²θ) and stays up longer (T grows with sin θ).
  5. The two angles are equally far from 45°, which is where the two paths meet as one.
Questions

Projectile at 60 degrees: quick answers

What is the maximum height of a ball thrown at 28 m/s at 60°?

With g = 9.8 m/s², H = 28² × sin²60° / (2 × 9.8) = 30 m.

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

T = 2 × 28 × sin 60° / 9.8 = 4.95 s and R = 28² × sin 120° / 9.8 = 69.28 m.

Which other angle gives the same range?

30°, because the two add to 90 degrees. It goes a third as high.

Try your own throw