# Projectile Motion

1. ### wowdusk

26
1. The problem statement, all variables and given/known data
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. Relevant 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?

2. ### LowlyPion

5,328
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))

3. ### wowdusk

26
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?

4. ### wowdusk

26
I think i solved this out...i got 61m???

5. ### LowlyPion

5,328
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.

6. ### wowdusk

26
Thank you...can you check if my answer is right?