Question:
A ball of mass 100 g is projected with velocity 20 $ms^{-1}$ at $60^{\circ}$ with horizontal. The decrease in kinetic energy of the ball during its entire upward journey is:
A ball of mass 100 g is projected with velocity 20 $ms^{-1}$ at $60^{\circ}$ with horizontal. The decrease in kinetic energy of the ball during its entire upward journey is:
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At max height, kinetic energy is due to horizontal velocity only.
Updated On: Jun 6, 2026
- 15 J
- 20 J
- Zero
- 5 J
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Projectile motion and energy.
Step 2: Meaning
Change in kinetic energy from projection to maximum height.
Step 3: Analysis
Initial $KE = \frac{1}{2}mv^2 = \frac{1}{2}(0.1)(20)^2 = 20~J$. At max height, only horizontal velocity remains: $v_x = v \cos(60^{\circ}) = 20(0.5) = 10~ms^{-1}$. $KE_{max_height} = \frac{1}{2}(0.1)(10)^2 = 5~J$. Decrease $= 20 - 5 = 15~J$.
Step 4: Conclusion
The kinetic energy decreases by 15 J.
Final Answer: (A)
Projectile motion and energy.
Step 2: Meaning
Change in kinetic energy from projection to maximum height.
Step 3: Analysis
Initial $KE = \frac{1}{2}mv^2 = \frac{1}{2}(0.1)(20)^2 = 20~J$. At max height, only horizontal velocity remains: $v_x = v \cos(60^{\circ}) = 20(0.5) = 10~ms^{-1}$. $KE_{max_height} = \frac{1}{2}(0.1)(10)^2 = 5~J$. Decrease $= 20 - 5 = 15~J$.
Step 4: Conclusion
The kinetic energy decreases by 15 J.
Final Answer: (A)
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