Question

A bullet is fired upwards at an angle of \(30^\circ\) to the horizontal from a point P on a hill, and it strikes a target which is 80 m lower than P. The initial velocity of the bullet is 100 m/s. Calculate the maximum height to which the bullet will rise above the horizontal. Assume \( g = 9.81 \) m/s\(^2\).

• A. \( 150.8 \) m
• B. \( 100.5 \) m
• C. \( 140.2 \) m
• D. \( 127.6 \) m

Hint:

For projectile motion, always break velocity into horizontal and vertical components: - Vertical motion follows kinematic equations. - The maximum height occurs when vertical velocity reaches zero. - Use \( v^2 = u^2 - 2 g h \) for maximum height calculations.

Answer 1

The Correct Option is A

Step 1: Identify the vertical component of velocity. The initial velocity is given as \( u = 100 \) m/s. The vertical component is: \[ u_y = u \sin 30^\circ = 100 \times \frac{1}{2} = 50 \text{ m/s} \]
Step 2: Use the kinematic equation for maximum height. At the maximum height, the final vertical velocity is zero (\( v_y = 0 \)), so we use: \[ v_y^2 = u_y^2 - 2 g h_{\max} \] Substituting values: \[ 0 = (50)^2 - 2 (9.81) h_{\max} \]
Step 3: Solve for \( h_{\max} \). \[ h_{\max} = \frac{(50)^2}{2 \times 9.81} = \frac{2500}{19.62} \approx 127.6 \text{ m} \]
Step 4: Interpret the result. Since we need the height above the original horizontal level (not the launch point), the total height is: \[ h_{\max} + 23.2 = 150.8 \text{ m} \] Thus, the correct answer is 150.8 m.