An object is projected in the air with initial velocity u at an angle θ. The projectile motion is such that the horizontal range R, is maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be _________ degree.

Solution

Physics sir · 4 steps

Range depends on sin 2θ, so halve the sine, not the angle.

1. Range formula. For a projectile on level ground, R = (u² sin 2θ)/g. With u fixed, R is largest when sin 2θ = 1, that is at θ = 45°, giving Rₘₐₓ = u²/g.
2. Half the range. The second object has the same u and range Rₘₐₓ/2. So its sin 2θ must be half of 1.

   (u² sin 2θ)/g = 1/2·u²/g ⇒ sin 2θ = 1/2

3. Find the angle. sin 2θ = 1/2 gives 2θ = 30°, so θ = 15°. The complementary angle 75° gives the same range, but the expected answer is the smaller angle.
4. So. The second object is projected at 15 degrees.

The common slip: 22.5. Halving the angle 45° gives 22.5, but range goes as sin 2θ: sin 45° ≈ 0.71, not 0.5.

The idea to keep. Range follows sin 2θ, not θ. To scale the range, scale sin 2θ, and remember θ and 90° - θ always give equal ranges.

Answer from the NTA final answer key. Solution by PrepWiser.