Motion in a Plane

45°

Step 1: The range formula is R = v₀² sin2θ / g. Step 2: For R to be maximum, sin2θ must be maximum. Step 3: Maximum value of sin2θ = 1, which occurs when 2θ = 90°, i.e., θ = 45°. At θ = 30° or 60°, sin2θ = sin60° or sin120° = √3/2 ≈ 0.866, which is less than 1. At θ = 90°, sin2θ = sin180° = 0 (no horizontal range).

3 m/s²

Step 1: Centripetal acceleration formula: a = v²/r. Step 2: v = 9 m/s, r = 27 m. Step 3: a = (9)² / 27 = 81/27 = 3 m/s². The acceleration is always directed towards the centre of the circle. Wrong options: 9 m/s² is v/r instead of v²/r; 27 m/s² is just the radius; 0.33 m/s² is r/v².

4000 N

Step 1: Centripetal force formula: F = mv²/r. Step 2: m = 1000 kg, v = 20 m/s, r = 100 m. Step 3: F = 1000 × (20)² / 100 = 1000 × 400 / 100 = 4000 N. Wrong options: 2000 N is mv/r (missing one power of v); 200 N is v²/r without mass; 8000 N is double the answer, a common arithmetic error.

v = √(rg tanθ)

Step 1: For a banked road without friction, resolve the normal force FN: horizontal component FN sinθ provides centripetal force, vertical component FN cosθ balances weight mg. Step 2: Dividing these two equations: FN sinθ / FN cosθ = (mv²/r) / mg, giving tanθ = v²/(rg). Step 3: Solving for v: v² = rg tanθ, so v = √(rg tanθ). The other options arise from incorrect resolution of forces.