anonymous
  • anonymous
Find the angle for which a projectile fired from a height h with velicity u at an angle theta will have max. range
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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experimentX
  • experimentX
0 of course ... to have max range you need to have max horizontal component ... vertical component to initial velocity will only make fall faster
anonymous
  • anonymous
0 is wrong. Give me a minute to elaborate.
experimentX
  • experimentX
what do you mean ... when fired from HEIGHT???

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anonymous
  • anonymous
It is fired for a height h upward with an angle theta
experimentX
  • experimentX
than that would be 45
anonymous
  • anonymous
Yeah and now we are firing from a height h
anonymous
  • anonymous
it would be different because the y component would be zero later than when fired from ground
anonymous
  • anonymous
Didn't see it was launched from a different height.
anonymous
  • anonymous
Yeah that's what makes the difference
experimentX
  • experimentX
|dw:1332870539943:dw||dw:1332870610564:dw|
anonymous
  • anonymous
I did not entirely get the point.Please elaborate
experimentX
  • experimentX
you meant first or second??
experimentX
  • experimentX
is your case??
anonymous
  • anonymous
I mean how can you prove that range would be max if you threw it horizontally.The second one
experimentX
  • experimentX
okay, i will prove it in terms of simple logic, let it be fired at angle theta, let the projectile flies to t seconds, the range will be u*cos theta* t it is maximum for value of theta = 0
anonymous
  • anonymous
Yeah that's what I thought at first.But the question phrased it differently.So I was just thinking if I missed something.
anonymous
  • anonymous
|dw:1332870995270:dw|
anonymous
  • anonymous
Sorry Im bad at drawing
anonymous
  • anonymous
Thats's what is supposed to be the answer.The question is asked to prove that it is the angle
experimentX
  • experimentX
t = ((sqrt(u^2*sinQ^2 + 2as) - u*sinQ))/a maximize this function R(Q) = ((sqrt(u^2*sinQ^2 + 2as) - u*sinQ))/a
anonymous
  • anonymous
pavankumartgpk Have you had calculus? This is much easier to solve using calculus.
anonymous
  • anonymous
Yeah Im pretty good with calculus.Go on
experimentX
  • experimentX
R(Q) = u*cosQ*((sqrt(u^2*sinQ^2 + 2as) - u*sinQ))/a
anonymous
  • anonymous
Is that the maximum range?
experimentX
  • experimentX
no function for range, ... you know how to find maxima right??
anonymous
  • anonymous
Yeah I do
anonymous
  • anonymous
So you are saying that I should maximise that function saying dr/dQ is 0?
experimentX
  • experimentX
yep
experimentX
  • experimentX
the graph of horizontal range vs Q will be like this .. Q on x-axis and Range on y-axis ... just check where range will be maximum
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anonymous
  • anonymous
Ok Ill try that.Just tell me how did you get the range function
experimentX
  • experimentX
the above function, and also i put 5 for initial velocity, 200 for distance
anonymous
  • anonymous
But the distance is what needs to be maximised
experimentX
  • experimentX
yes, range is the distance ... which is a function of angle, maximize distance for values of Q, i think it is quite clear from above graph, you range tends to zero as Q increase to 90 and greatest when it is zero
anonymous
  • anonymous
|dw:1332873329556:dw| But how can you say that is the graph.I mean the answer can't be 0
anonymous
  • anonymous
Realize that at the point of landing in the vertical is expressed as\[0= h + v_y t - 0.5 g t^2\]and in the horizontal as\[R = v_x t\]Solve the second to t and substitute into the first\[0 = h + {v_y R \over v_x} - 0.5 g {R^2 \over v_x^2}\]Realize that\[v_y = v \sin(\theta) ~ {\rm and}~ v_x = v \cos(\theta)\]Therefore\[0 = h + {\sin(\theta) \over \cos(\theta)} R - 0.5 g {R^2 \over v^2 \cos^2(\theta)}\]Multiply through by \(\cos^2(\theta)\)\[0 = h \cos^2(\theta) + \sin(\theta)\cos(\theta)R - 0.5g {R^2 \over v^2}\]Multiply through by -2\[2h \cos^2(\theta) = -\sin(2\theta) R + g {R^2 \over v^2}\](I made the substitution that sin(2x) = 2sin(x)cos(x)). Let's substitute \[\cos(2\theta) = 2\cos^2(\theta) - 1\]\[h(1+\cos(2\theta)) = -\sin(2\theta) R + g{R^2 \over v^2}\]If we differentiate this expression w.r.t. \(\theta\) and set \({dR \over d\theta} = 0\) We can come up with the following expression for \(\theta\)\[\theta = \arccos \sqrt{2 gh + v^2 \over 2gh + 2 v^2}\] You can see a more detailed derivation here : http://en.wikipedia.org/wiki/Range_of_a_projectile#Maximum_range or here https://docs.google.com/viewer?a=v&q=cache:v4OJ5fWgeloJ:usna.edu/Users/physics/mungan/Scholarship/Projectile.pdf+maximum+range+when+projectile+fired+from+some+height&hl=en&gl=us&pid=bl&srcid=ADGEEShV6NgyIvr7Ib0uJRAKD_9PSgaxgCxUN1uMwiXU0he-OGeY-dHLIGYHiuprgrNGqSF9NOw9Vb-YVVq6_VY6Zf73Flx1S6LnUOymRyGNUBCc8QxWiEedJn-beAZSVLejnrpop2JL&sig=AHIEtbSidjOTl-jNrUk2lALgmJu38vQ62A
anonymous
  • anonymous
Because if it is indeed 0 the time of flight in the y axis would be lesser that when fired at an angle.So if it is fired at an angle it will start with some velocity in the y direction increasing the time of flight which inturn increases the range
experimentX
  • experimentX
Oo... i think i get the picture now, it was projected above
experimentX
  • experimentX
from certain height ....
anonymous
  • anonymous
Yeah
anonymous
  • anonymous
Ok thanx guys
experimentX
  • experimentX
damn this is taking too long, |dw:1332874202452:dw| i guess you mean this
anonymous
  • anonymous
Yep!Exactly
experimentX
  • experimentX
did you get the answer ??
anonymous
  • anonymous
Yeah I did.I differentiated the range function taking dR/dQ=0 and then I got it
experimentX
  • experimentX
okay then that solves it.
anonymous
  • anonymous
And of course also taking y=0
anonymous
  • anonymous
Yeah it does.Thanx

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