Question

A cannon placed on a cliff at a height of 375 m fires a cannonball with a velocity of 100 m/s\(^{-1}\) at an angle of 30° above the horizontal. The horizontal distance between the cannon and the target is:
(Acceleration due to gravity \( g = 10 \) ms\(^{-2}\))

• A. \( 750\sqrt{3} \) m
• B. \( 500\sqrt{3} \) m
• C. \( 250\sqrt{3} \) m
• D. \( 750 \) m

Hint:

Break motion into horizontal and vertical components. - Use time of flight to compute horizontal range.

Answer 1

The Correct Option is A

Step 1: Compute time of flight
Vertical velocity: \[ u_y = 100 \sin 30° = 50 \text{ m/s}. \] Solving for time of flight \( T \): \[ T = \frac{u_y + \sqrt{u_y^2 + 2gh}}{g}. \] \[ T = \frac{50 + \sqrt{50^2 + 2(10)(375)}}{10}. \] Step 2: Compute horizontal range
Horizontal velocity: \[ u_x = 100 \cos 30° = 50\sqrt{3} \text{ m/s}. \] \[ R = u_x T = 50\sqrt{3} \times 15 = 750\sqrt{3}. \] Thus, the correct answer is \( \boxed{750\sqrt{3}} \) m.