A) \[E\]
B) \[\frac{E}{2}\]
C) \[\frac{E}{\sqrt{2}}\]
D) \[0\]
Correct Answer: B
Solution :
Given: Kinetic energy of the ball = E and angle of projection\[(\theta )=45{}^\circ \]. We know that velocity of the ball at the highest point\[=v\cos \theta =v\cos 45{}^\circ =\frac{v}{\sqrt{2}}.\] Therefore kinetic energy of the ball\[=\frac{1}{2}m\times {{\left( \frac{v}{\sqrt{2}} \right)}^{2}}\]\[=\frac{1}{4}m{{v}^{2}}=\frac{E}{2}.\]
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