Question:
A cricket ball is hit at an angle of 30° to the horizontal with a kinetic energy E. Its kinetic energy when it reaches the highest point is
A cricket ball is hit at an angle of 30° to the horizontal with a kinetic energy E. Its kinetic energy when it reaches the highest point is
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At highest point, only horizontal velocity contributes to kinetic energy.
Updated On: May 8, 2026
- $\frac{E}{2}$
- 0
- $\frac{2E}{3}$
- $\frac{3E}{4}$
- E
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The Correct Option is D
Solution and Explanation
Concept:
At the highest point of projectile motion, vertical velocity becomes zero, but horizontal velocity remains unchanged. Hence kinetic energy is due to horizontal component only.
Step 1: Initial kinetic energy. \[ E = \frac{1}{2}mv^2 \]
Step 2: Resolve velocity. \[ v_x = v\cos 30^\circ = \frac{\sqrt{3}}{2}v \]
Step 3: Velocity at highest point. Only horizontal component remains: \[ v = \frac{\sqrt{3}}{2}v \]
Step 4: Kinetic energy at top. \[ E' = \frac{1}{2}m\left(\frac{\sqrt{3}}{2}v\right)^2 = \frac{1}{2}m \cdot \frac{3}{4}v^2 = \frac{3}{4}E \]
Step 5: Final answer. \[ \boxed{\frac{3E}{4}} \]
Step 1: Initial kinetic energy. \[ E = \frac{1}{2}mv^2 \]
Step 2: Resolve velocity. \[ v_x = v\cos 30^\circ = \frac{\sqrt{3}}{2}v \]
Step 3: Velocity at highest point. Only horizontal component remains: \[ v = \frac{\sqrt{3}}{2}v \]
Step 4: Kinetic energy at top. \[ E' = \frac{1}{2}m\left(\frac{\sqrt{3}}{2}v\right)^2 = \frac{1}{2}m \cdot \frac{3}{4}v^2 = \frac{3}{4}E \]
Step 5: Final answer. \[ \boxed{\frac{3E}{4}} \]
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