A) E
B) \[E/\sqrt{2}\]
C) \[E/2\]
D) zero
Correct Answer: C
Solution :
[c] Let u be the speed with which the ball of mass m is projected. Then the kinetic energy (E) at the point of projection is \[E=\frac{1}{2}m{{u}^{2}}\] ?(i) When the ball is at the highest point of its flight, the speed of the ball is \[\frac{u}{\sqrt{2}}\] (Remember that the horizontal component of velocity does not change during a projectile motion). \[\therefore \]The kinetic energy at the highest point \[=\frac{1}{2}m{{\left( \frac{u}{\sqrt{2}} \right)}^{2}}=\frac{1}{2}\frac{m{{u}^{2}}}{2}=\frac{E}{2}\][From (i)]You need to login to perform this action.
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