Question #40610

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Question #40610

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Answer 1 · 2017-01-21T08:38:47

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As referred to the figure above.

Kinetic energy at point of projection $K E_{projection} = \frac{1}{2} m u^{2}$ $KE_"projection"=1/2m u^2$
Maximum height of the object having mass $m$ $m$ will be reached when $sin θ$ $sin theta$ component of projectile's velocity becomes $0$ $0$ (zero).

As such kinetic energy at this point is given by
$K E_{max height} = \frac{1}{2} m {(u cos θ)}^{2}$ $KE_"max height"=1/2m(ucos theta)^2$
Ratio of the two is $\frac{K E_{max height}}{K E_{projection}} = \frac{\frac{1}{2} m {(u cos θ)}^{2}}{\frac{1}{2} m u^{2}}$ $(KE_"max height")/(KE_"projection")=(1/2m(ucos theta)^2)/(1/2m u^2)$
$= {cos}^{2} θ$ $=cos^2 theta$

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Question #40610

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