Projectile launch point problem

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Homework Help Overview

The discussion revolves around a projectile motion problem involving a projectile launched at a specific speed and angle, with a focus on determining the launch point's height relative to the landing point and the maximum height reached by the projectile.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations to find the launch point's height and the maximum height of the projectile. Some participants suggest considering the vertical component of velocity at maximum height and provide hints regarding relevant equations.

Discussion Status

Participants have offered hints and guidance on how to approach the problem, particularly regarding the conditions at maximum height. The original poster has indicated some progress in understanding but still seeks clarity on setting up the equations correctly.

Contextual Notes

The original poster notes confusion regarding the use of specific values related to the landing point when calculating the maximum height, which introduces uncertainty in their approach.

SA32
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I'm having some difficulty with a projectiles question.

"A projectile is fired with an initial speed of 26.0 m/s at an angle of 40.0 degrees above the horizontal. The object hits the ground 8.00 s later."

The first part of the question is: "How much higher or lower is the launch point relative to the point where the projectile hits the ground?
Express a launch point that is lower than the point where the projectile hits the ground as a negative number."

Which I already found out using y=(viy)(t)-(1/2)(g)(t^2)

Substituting the given numbers, I get -180 m, which would be the case if the landing point was below the launch point. Since the question wants a launch point lower than the landing point, the answer is 180 m.

The second part, where I'm having trouble: "To what maximum height above the launch point does the projectile rise?"

What I have,

x=(26)(cos(40))(t)
y=(26)(sin(40))(t)-(1/2)(g)(t^2)

I can't substitute 180 m or 8 s because those numbers apply to the landing point, and I'm looking for ymax so I end up with three unknowns in two equations, which I cannot solve. I don't know how to set this up so that I can solve it.

Thanks for any help!
 
Last edited:
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To find maximum height, remember that the vertical component of velocity at the maximum height is zero.

Does this help?
 
Hint: Try using the formulae;

[tex]v_{f}^{2}=v_{i}^{2}+2ax[/tex]

:wink:

Edit: Beaten to it.
 
Whoops! Should post here to let you know I used your hints to figure it out, so thanks!
 

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