Question

A ball is thrown from the ground to clear a wall 3 m high at a distance of 6 m and falls 18 m away from the wall, the angle of projection of ball is

• A. \( \tan^{-1}\left( \frac{3}{2} \right) \)
• B. \( \tan^{-1}\left( \frac{2}{3} \right) \)
• C. \( \tan^{-1}\left( \frac{1}{2} \right) \)
• D. \( \tan^{-1}\left( \frac{3}{4} \right) \)

Hint:

For projectile motion, use the equations of range and height to solve for the angle of projection.

Answer 1

The Correct Option is B

Using the equations of projectile motion, we can determine the angle of projection. The angle of projection \( \theta \) can be calculated using the relation for the range and height of the projectile. The correct angle is \( \tan^{-1}\left( \frac{2}{3} \right) \).