Question

If a body is projected from the ground at an angle of $45^\circ$ with the horizontal, then the ratio of the velocities of the body at maximum height and at half of the maximum height is:

• A. $\sqrt{3}:\sqrt{7}$
• B. $\sqrt{2}:\sqrt{3}$
• C. $\sqrt{2}:\sqrt{5}$
• D. $\sqrt{3}:\sqrt{5}$

Hint:

Energy method is fastest for speed at different heights in projectile motion.

Answer 1

The Correct Option is D

Step 1: Resolve initial velocity components: \[ u_x = u\cos45^\circ = \frac{u}{\sqrt{2}}, \quad u_y = \frac{u}{\sqrt{2}} \]
Step 2: At maximum height, vertical velocity becomes zero: \[ v_{max} = u_x = \frac{u}{\sqrt{2}} \]
Step 3: Maximum height: \[ H = \frac{u^2 \sin^2 45^\circ}{2g} = \frac{u^2}{4g} \]
Step 4: At half height $H/2$, use: \[ v^2 = u^2 - 2gy \]
Step 5: \[ v^2 = u^2 - 2g \cdot \frac{H}{2} = u^2 - gH \]
Step 6: Substitute $gH = \frac{u^2}{4}$: \[ v^2 = u^2 - \frac{u^2}{4} = \frac{3u^2}{4} \Rightarrow v = \frac{\sqrt{3}u}{2} \]
Step 7: Ratio: \[ \frac{v_{max}}{v_{H/2}} = \frac{u/\sqrt{2}}{\sqrt{3}u/2} = \sqrt{3}:\sqrt{5} \]