If the kinetic energy at maximum height is 2/5 the kinetic energy at half the maximum height, find the angle of projection.

In other words: K.E. at H(max) = 0.4*K.E. at 0.5*H(max)

I got 60 degrees, however, my teacher told me that the answer is 30 degrees. Please solve it for me so I can see where I went wrong.

Well here's my edit:

Consider u to be the velocity at H(max) Consider v to be the velocity at half of H(max) Consider c to be initial velocity given Consider @ to be angle projected at sqr() means square root

u^2 = 0.4v^2

u is obviously c*cos(@) v I found to be: sqr((0.5u^2)(1+(cos(@))^2)

Using this, I substituted into the previous equation and solved to attain: cos@ = 0.5

So.... That gives me 60 degrees. Am I right? Or is there some sort of catch?

  • 2
    $\begingroup$ No, show your solution first, so that we can see where you went wrong. $\endgroup$ – Bernhard Jun 13 '14 at 9:29
  • $\begingroup$ @Gummy you are right i also got 60 degree. $\endgroup$ – Rahul kumar walia Jun 13 '14 at 9:43
  • $\begingroup$ Please read the site's homework policy and act accordingly. That's the point that @Bernhard is trying to make. $\endgroup$ – 299792458 Jun 13 '14 at 10:17
  • $\begingroup$ If you'd really act according to the home-work policy of this site then perhaps you will have to delete the question. Because acc. to their policy one can only ask about the concept not how to solve this. $\endgroup$ – user31782 Jun 13 '14 at 11:14
  • $\begingroup$ @user31782 I understand you are pissed for some reason. I have also been a victim of ''silent downvotes'' (which you keep ''countering'') in one of my answers, but I still think that the homework policy of the site is rational. God bless you, kid, in terms of the proposal that I saw on your flair, even though I'm precisely the kind of a person who would stand eliminated by your mission statement. $\endgroup$ – 299792458 Jun 13 '14 at 15:49

Velocity at top point is totally horizontal,$ v\cos x$ velocity at half the top height is $V=\sqrt{ {(v\sin x-gt)}^{2}-{(v\cos x)}^{2} } $ where $x$ is the angle of projection $g$ is acceleration due to gravity $t$ is time elapsed at half of max height $v$ is the velocity of projection

Equate ${v\cos x}^{2}= 2/5* {V}^{2} $


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