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JEE MAIN Physics QUESTION #992
Question 1
A ball of mass 100 g is projected with velocity \(20\ \text{m/s}\) at \(60°\) with horizontal. The decrease in kinetic energy from projection to highest point is:
  • 5 J
  • 15 J✔️
  • 20 J
  • zero
Correct Answer Explanation
At highest point, only horizontal component remains: \(v_x = 20\cos60° = 10\ \text{m/s}\). Initial KE \(= \frac{1}{2}(0.1)(400)=20\ \text{J}\). Final KE \(= \frac{1}{2}(0.1)(100)=5\ \text{J}\). Decrease \(= 20-5=15\ \text{J}\).