Numford
Projectile motion

Why the range is maximum at 45 degrees

sin 2θ is largest at θ = 45°, so the range is largest there: R max = u² / g = 40 m at 20 m/s with g = 10 m/s².

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 T2.83 s2u sin θ / g
Range R40 mu² sin 2θ / g
Max height H10 mu² sin²θ / 2g
Velocity at 0.71 s15.81 m/svx = 14.14, vy = 7.07, 26.57° up
Pathy = x - 0.025x²y = x tan θ - gx² / (2u² cos²θ)

A ball thrown at 20 m/s and 45 degrees rises 10 m, flies for 2.83 s and lands 40 m away.

g (m/s²)
Show the formulas
Time of flight
T = 2u sin θ / g = 2 × 20 × sin 45° / 10 = 2.83 s
Range
R = u² sin 2θ / g = 20² × sin 90° / 10 = 40 m
Max height
H = u² sin²θ / 2g = 20² × sin²45° / (2 × 10) = 10 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 = 14.14 m/s, vy = u sin θ - gt = 7.07 m/s at t = 0.71 s, so v = 15.81 m/s
Path
y = x tan θ - gx² / (2u² cos²θ), here y = x - 0.025x²
Working

Step by step

  1. Given u = 20 m/s at θ = 45°, with g = 10 m/s².
  2. R = u² sin 2θ / g, and sin 2θ is largest, 1, when 2θ = 90°, so θ = 45°.
  3. R max = u² / g = 20² / 10 = 40 m
  4. H = u² sin²θ / 2g = 20² × sin²45° / (2 × 10) = 10 m
  5. T = 2u sin θ / g = 2 × 20 × sin 45° / 10 = 2.83 s
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

  1. The range is R = u² sin 2θ / g, and only sin 2θ changes with the angle.
  2. The largest value of a sine is 1, at 90°. So 2θ = 90° and θ = 45°.
  3. The longest range is then R max = u² / g.
  4. At 45° the height is H = u² / 4g, a quarter of the range.
  5. With air resistance the best angle is a little lower than 45°, but in NCERT problems air is ignored.
Questions

Maximum range at 45 degrees: quick answers

Why is the range maximum at 45 degrees?

R = u² sin 2θ / g, and sin 2θ reaches its largest value, 1, when 2θ = 90 degrees, that is θ = 45 degrees.

What is the maximum range at 20 m/s?

R max = u² / g = 20² / 10 = 40 m, with H = 10 m and T = 2.83 s.

How is the height related to the range at 45 degrees?

At 45 degrees H = u² / 4g, exactly a quarter of the range.

Try your own throw