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Projectile Motion and maximum height

  • Thread starter wowdusk
  • Start date
1. Homework Statement
A projectile is launched with a speed of 40 m/s at an angle of 60 degrees above the horizontal. Use conservation of energy to find the maximum height reached by the projectile during its flight.

2. Homework Equations
KEi+PEi=KEf+PEf


3. The Attempt at a Solution
i am not sure where vf would come from. Would it be just Vi*cos(60)?
I am not sure why that makes sense.
Does Vix=Vf?...and why?
 

LowlyPion

Homework Helper
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1. Homework Statement
A projectile is launched with a speed of 40 m/s at an angle of 60 degrees above the horizontal. Use conservation of energy to find the maximum height reached by the projectile during its flight.

2. Homework Equations
KEi+PEi=KEf+PEf


3. The Attempt at a Solution
i am not sure where vf would come from. Would it be just Vi*cos(60)?
I am not sure why that makes sense.
Does Vix=Vf?...and why?
Assume your potential energy is 0 when you launch and your kinetic energy in the y direction is what? 1/2*m*Viy2?

And at the height it has no vertical kinetic energy and the potential energy is what?

What is the vertical component of V? (Hint: it's not Vi*cos(60))
 
why do i need the vertical component of Vi

I dont know how to find the V at the heighest point...

At the heighest point is the V in vertical direction 0 anyway?
 
I think i solved this out...i got 61m???
 

LowlyPion

Homework Helper
3,079
4
why do i need the vertical component of Vi

I dont know how to find the V at the heighest point...

At the heighest point is the V in vertical direction 0 anyway?
Yes your final V is 0.

But your initial V is the vertical component of V, as that is the component of V that is affected by gravity ... you know, where that potential energy is building.
 
Thank you...can you check if my answer is right?
 

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