Q61 / 75 · JEE Main 2026 (22 January Shift 1) · Physics

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

  • Correct answer
    A
    $20 \sqrt{2}$
  • B
    40
  • C
    $40 \sqrt{2}$
  • D
    $20 \sqrt{3}$

Solution

Horizontal velocity is constant: $v_x = u\cos 60° = \frac{u}{2}$.
When direction is $45°$ with horizontal and speed is 20 m/s:
$v_x = 20\cos 45° = \frac{20}{\sqrt{2}}$.
$\frac{u}{2} = \frac{20}{\sqrt{2}} \Rightarrow u = \frac{40}{\sqrt{2}} = 20\sqrt{2}$ m/s.

Hence, the correct answer is option: 'A'.

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