Difficult projectile problem

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  • #1
ozone
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A projectile is fired with an initial velocity v such that it passes through two points that are both a distance h above the horizontal. Show that if the gun is adjusted for maximum range the separation of the two points is

[itex] d = v/g \sqrt{v^2 - 4gh} [/itex]


Homework Equations



I have been struggling at this problem for much of the day. I even derived the original equations for range and max height

range, height [itex] R = v^2/g , H = v^2 / 4g [/itex]
Also we have our conservation of motion equations, and the distance function of a projectile (under the constraints of max range)
[itex] F(t) = (1/\sqrt{2} vt) \overline{i} + (1/\sqrt{2} vt - 1/2 gt^2) \overline{j} [/itex]

The Attempt at a Solution


I attempted to use conservation of energy to find a solution to this problem, but I was having a lot of difficulty relating it to my distance function. I also tried using my distance function but I found that I could not effectively remove the time variable.
 
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Answers and Replies

  • #2
rl.bhat
Homework Helper
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Use the formula for projectile which does not contain t.
y = x tan(theta) - g x^2sec^(theta)/2v^2
Put y = h and solve for x. Difference between x gives you the distance d.
 
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  • #3
ozone
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Thanks.. putting y in terms of x made this problem quite easy to solve. I wish I had seen it myself.
 
  • #4
voko
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Another way to solve it would be to observe that since the angle is 45, you effectively know the vertical and horizontal components of the velocity. So you can immediately compute time t to height h. The equation involved is quadratic, so it gives you two roots. Then the distance sought is the difference between the roots multiplied by the horizontal speed.
 
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