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Max horizontal distance wrt variable angle

  • #1

Homework Statement


A ball is thrown with speed v from height h. What angle should it be thrown so it travels the maximum horizontal distance?

I'll assume the angle theta is the angle from the horizontal to the initial direction of the velocity.

Okay, since there is no air resistance we know:
[tex]x = x_{0} + v_{0x}t + \frac{1}{2}a_{x}t^{2}[/tex]
since there is no acceleration in that direction, the distance the ball can reach would be:
[tex]x = v_{0}cos(\theta)t[/tex]

Since, as it stands, x is a function of both t and theta I will use the vertical motion to find the time of flight as a function of theta and plug that into the above equation.

I think I have to do this in two parts (or maybe I've been working too hard and have confused myself).

Part1: from launch to maximum height (ie v_y = 0)
Part2: from maximum height to landing

Part1:
[tex]y = h + v_{0}sin(\theta)t - \frac{1}{2}gt^{2}[/tex]
We don't know what the maximum height is but we can take the derivative to find at what time it occurs (and thus find it).
[tex]v_{y} = v_{0}sin(\theta) - gt[/tex]
[tex]t_{1} = \frac{v_{0}sin(\theta)}{g}[/tex]

Plugging this into the first equation of Part 1 yields:
[tex]y_{max} = h + \frac{1}{2g}v_{0}^{2}sin(\theta)^{2}[/tex]

Part2:
[tex]y = y_{max} + v_{0}t - \frac{1}{2}gt^{2}[/tex]
here v_0 is zero and y_final = 0
[tex]y_{max} = \frac{1}{2}gt^{2}[/tex]
after some simplifying:
[tex]t_{2} = \frac{1}{g}\sqrt{2gh + v_{0}sin(\theta)^{2}}[/tex]

So the total time of flight would be:
[tex]t = t_{1} + t_{2} = \frac{v_{0}sin(\theta) + \sqrt{2gh + v_{0}^{2}sin(\theta)^{2}}}{g}[/tex]

plugging this into the equation for horizontal distance:
[tex]x = v_{0}cos(\theta)(\frac{v_{0}sin(\theta) + \sqrt{2gh + v_{0}^{2}sin(\theta)^{2}}}{g})[/tex]

now, as a check, letting h = 0 we would get:
[tex]x = \frac{2v_{0}^{2}}{g}cos(\theta)sin(\theta)[/tex]
which gives an angle of 45 degrees being optimal (as kinda expected).

Now, without that simplification that above expression is nasty and I can't find a way to simplify it or its derivative which leads me to believe I messed up earlier >.<
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
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hi alexa! :smile:

(have a theta: θ and a square-root: √ and try using the X2 and X2 icons just above the Reply box :wink:)
I think I have to do this in two parts (or maybe I've been working too hard and have confused myself).

Part1: from launch to maximum height (ie v_y = 0)
Part2: from maximum height to landing
sorry, but yes you've been working too hard …

get some sleep! :zzz: …

then just go straight to solving the original equation to find t when y = 0 :wink:
 
  • #3
Ha... maybe I could use a nap...

But I recover the same travel time (as I should) and I'm still left with something that I can't simplify.
 
  • #4
tiny-tim
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show us your solution for y = 0, so that we can see what's going wrong :wink:
 
  • #5
okay, I start here:
[tex]y = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}[/tex]
we know y = 0
y_0 = h
v_0y = v_0 sinθ
[tex]0 = h + v_{0}sin(\theta)t - \frac{1}{2}gt^{2}[/tex]
I'll use the quadratic formula to find the roots:

[tex]t = \frac{1}{g}(v_{0}sin(\theta) + \sqrt(v_{0}^{2}sin{2}(\theta) + 2gh)[/tex]
 
  • #6
tiny-tim
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ok, now multiply by cosθ and find the maximum :wink:
 
  • #7
well thats my issue! To find the maximum I'd have to take a derivative wrt theta, no?

[tex]x = \frac{1}{g}(v_{0}^{2}sin(\theta)cos(\theta) + v_{0}cos(\theta)\sqrt{2gh + v_{0}^{2}sin^{2}(\theta)}[/tex]
 
  • #8
tiny-tim
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yes! :rolleyes:

do it!!
 
  • #9
[tex]\frac{v_{0}^{2}}{g}cos(2x) + \frac{vsin(\theta)(v^{2}cos(2\theta) - 2gh)}{g\sqrt{2gh + v^{2}sin^{2}(\theta)}}[/tex]

Setting it equal to zero and doing some simplifying:

[tex]cos(2\theta)\sqrt{2gh + v^{2}sin^{2}(\theta)} = 2ghsin(\theta) - v^{2}cos(2\theta)sin(\theta)[/tex]

more simplifying doesn't seem to get me anywhere.... although I just missed a sin in there, let me see what I can do now...
 

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