A ball whose kinetic energy is ( E = 1 2 mu ^ 2 ) , is projected at an angle of 45° to…
Physics · JEE Main · NTA Exams — Kinematics
A ball whose kinetic energy is \(\left(\mathrm{E}=\frac{1}{2} \mathrm{mu}^{2}\right)\), is projected at an angle of 45° to the horizontal. The kinetic energy of the ball at the highest point of its flight will be
- E
- \(\frac{\mathrm{E}}{\sqrt{2}}\)
- \(\frac{E}{2}\)
- zero
Answer
(C) E 2
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