Projectiles A and B are thrown at angles of 45° and 60° with vertical respectively from top of a 400 m high tower. If their ranges and times of flight are same, the ratio of their speeds of projection vA: vB is: [Take g = 10 m s⁻² ] (a) 1: 2 (b) √(2): 1 (c) 1: √(2) (d) 1: √(3)

Answer: (b). From the same height, equal times of flight need equal vertical components: vAcos 45° = vBcos 60° (angles from vertical). (vA)/(vB) = (cos 60°)/(cos 45°) = (1/2)/(1/√(2)) = 1/√(2) Equal ranges with equal times would also need vAsin 45° = vBsin 60°, i.e. √(3): √(2), so both conditions cannot hold together. Using the time-of-flight condition, vA: vB = 1: √(2), option (c).

Projectiles A and B are thrown at angles of 45° and 60° with vertical… | JEE Main 2024 (30 Jan, Shift 2)

Q10JEE Main 2024 · 30 JAN · SHIFT 2Difficult

Comparing Two Projectiles

Projectiles A and B are thrown at angles of $45^\circ$ and $60^\circ$ with vertical respectively from top of a $400\text{ m}$ high tower. If their ranges and times of flight are same, the ratio of their speeds of projection $v_A : v_B$ is:

[Take $g = 10\text{ m s}^{-2}$ ]

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Answer: (b)

From the same height, equal times of flight need equal vertical components: $v_A\cos 45^\circ = v_B\cos 60^\circ$ (angles from vertical).
$\frac{v_A}{v_B} = \frac{\cos 60^\circ}{\cos 45^\circ} = \frac{1/2}{1/\sqrt{2}} = \frac{1}{\sqrt{2}}$
Equal ranges with equal times would also need $v_A\sin 45^\circ = v_B\sin 60^\circ$ , i.e. $\sqrt{3} : \sqrt{2}$ , so both conditions cannot hold together.
Using the time-of-flight condition, $v_A : v_B = 1 : \sqrt{2}$ , option (c).

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