Question

One second after projection, a projectile is travelling in a direction inclined at \( 45^\circ \) to horizontal. After two more seconds it is travelling horizontally. Then the magnitude of velocity of the projectile is ( \( g = 10 \) ms\(^{-2}\)):

• A. \( 10\sqrt{13} \) ms\(^{-1} \)
• B. \( 11 \) ms\(^{-1} \)
• C. \( 10\sqrt{2} \) ms\(^{-1} \)
• D. \( 20 \) ms\(^{-1} \)

Hint:

Breaking velocity components into horizontal and vertical parts simplifies projectile motion calculations.

Answer 1

The Correct Option is A

Step 1: Analyze vertical and horizontal velocity components. Given that the projectile moves at \( 45^\circ \) after one second, we use: \[ v_y = u \sin\theta - gt \] After one second, \( v_y = u \cos\theta \). Solving, we find \( u = 10\sqrt{13} \).