Projectile motion and release angle

  • Thread starter BananaRed
  • Start date
  • #1
7
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Determine the release angle that gives the maximum range for a projectile whose launch height and landing height are different. :confused: :confused: :confused: :confused:
 

Answers and Replies

  • #2
Need to be more specific. What's the landscape look like?

cookiemonster
 
  • #3
7
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that's all I was given
 
  • #4
52
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well first thing, I'm guessing your instructor wants you to come with an equation.

But first things first. When you release the ball, the point in the trajectory where it reaches the level it lands on, the ball will have to be traveling at 45 degrees. So at the level of the landing point, your horizontal and vertical velocities must be equal. It's easy to visualize when throwing at a level higher than your initial, but harder when it is lower.

For the solution, think position equation, and vectors.

So first we have [tex] \sin(\theta) v - \cos(\theta) v = a t [/tex]

Solving t for position equation ( [tex] s(t) = \frac{1}{2}at^2 + vt + s [/tex] ) using the quadratic equation, 1/2 acceleration cancels and we're left with

[tex] sin(\theta) v - \cos(\theta) v = \frac{-v \pm \sqrt{v^2 - 4 \frac{a}{2} s}}{1} [/tex]

I'll let you do the rest, or else it won't be your work.

And someone plz correct me if I'm wrong.
 
Last edited:

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