A projectile is thrown upward at an angle 60° with the horizontal. The speed of the projectile is 20 m/s when its direction of motion is 45° with the horizontal. The initial speed of the projectile is m/s. (a) 20√(3) (b) 20√(2) (c) 40 (d) 40√(2)

Answer: (b). Horizontal component is constant: ucos 60° = vcos 45° u × 1/2 = 20 × 1/√(2) = 10√(2) u = 20√(2) m/s So the answer is (b) 20√(2).

A projectile is thrown upward at an angle 60° with the horizontal… | JEE Main 2026 (22 Jan, Shift 1)

Q3JEE Main 2026 · 22 JAN · SHIFT 1Medium

Velocity, Momentum & Energy in Flight

A projectile is thrown upward at an angle $60^\circ$ with the horizontal. The speed of the projectile is $20\text{ m/s}$ when its direction of motion is $45^\circ$ with the horizontal. The initial speed of the projectile is $\underline{\qquad}\text{ m/s}$ .

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Answer: (b)

Horizontal component is constant: $u\cos 60^\circ = v\cos 45^\circ$
$u \times \frac{1}{2} = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2}$
$u = 20\sqrt{2}\text{ m/s}$
So the answer is (b) $20\sqrt{2}$ .

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