Kinematics

JEE Main level, single correct, medium

204 of 308

Question 204 of 308

A projectile is thrown upward at an angle 60∘60^{\circ} with the horizontal. The speed of the projectile is 20 m/s20 \mathrm{~m} / \mathrm{s} when its direction of motion is 45∘45^{\circ} with the horizontal. The initial speed of the projectile is ____\_\_\_\_ m/s\mathrm{m} / \mathrm{s}.

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Answer: (A) 20220 \sqrt{2}

Horizontal velocity is constant: vx=ucos⁡60°=u2v_x = u\cos 60° = \frac{u}{2}.
When direction is 45°45° with horizontal and speed is 20 m/s:
vx=20cos⁡45°=202v_x = 20\cos 45° = \frac{20}{\sqrt{2}}.
u2=202⇒u=402=202\frac{u}{2} = \frac{20}{\sqrt{2}} \Rightarrow u = \frac{40}{\sqrt{2}} = 20\sqrt{2} m/s.

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