TS EAMCET · Physics · Work Power Energy
A particle having kinetic energy \(K\) is projected at \(60^{\circ}\) with the horizontal. The kinetic energy at the highest point is
- A \(K\)
- B zero
- C \(K / 4\)
- D \(K / 2\)
Answer & Solution
Correct Answer
(C) \(K / 4\)
Step-by-step Solution
Detailed explanation
The motion of projectile is as shown in figure, Let initial velocity of projection is \(u\), so initial kinetic energy is \(\mathrm{KE}_1=\frac{1}{2} m u^2\) At highest point, the vertical component of velocity is zero, while the horizontal component is \(u \cos \theta\), which…
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