Exploding Projectiles

  • Thread starter jessicak
  • Start date
  • #1
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Homework Statement



A projectile is launched with speed v0 at an angle [tex]\theta[/tex] with respect to the horizontal. At the peak of its motion, it explodes into two pieces of equal mass, which continue to move in the original plane of motion. One piece strikes the ground a horizontal distance D farther from the launch point than the point directly below the explosion at a time t<v0sin[tex]\theta[/tex]/g. How high does the other piece go? Answer in terms of v0, [tex]\theta[/tex], g, and t.

Homework Equations


m1v1=m2v2
h=v02sin2([tex]\theta[/tex]/g


The Attempt at a Solution


I know I should be doing a momentum balance in the vertical direction before and after the collision, but I'm struggling to get the right equation. Any help would be appreciated
 

Answers and Replies

  • #2
IBY
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@jessicak
The way I see it is this: Firstly, focus on the fact that they both have equal mass, and the fact that at the peak of its motion, vertical velocity is zero, but the horizontal velocity keeps going. Look at conservation of momentum and see what happens. Secondly, the kinematic equation for both velocities have two components. One for horizontal, and the other vertical. Now, for the first piece, find the time it takes for it too hit the ground. Remember, the forward component and vertical components are separate, so only one of them decides the time it takes, then the other one decides the distance from explosion point.
 
  • #3
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Hmmm... I'm still very lost. If I take into consideration the time it takes for the first piece to reach the ground, how does that help me find the height of the second piece?
 
  • #4
IBY
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Find the kinematic equation of the second piece right when it explodes. And the best part is that since you are only finding the height, for this one, you don't have to take account of the horizontal velocity.
 
  • #5
26
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The general kinematic equation for height of a projectile is
h= v02sin2[tex]\theta[/tex]/g

How can I adjust this for the explosion? (I think time fits in there somehow)
 
  • #6
IBY
106
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Why is there no time component for that kinematic equation? Either that, or it is probably at maximum height at which the explosion happens.
 
  • #7
IBY
106
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BTW, the way I derived that equation, I got:

[tex]h=\frac{v^2sin^2\theta}{2g}[/tex]

Are you sure there is no two in the denominator?
 
  • #8
559
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I'm pretty sure it's just g in the denominator if I'm remembering correctly.
 
  • #9
IBY
106
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@iRaid
Yep, you are right, I got clumsy in one of the steps. :)
 
  • #10
26
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Unfortunately, my online homework does not accept that answer as correct. However, I am confident that time needs to be present somewhere, as the question distinctly asks for it.
 

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