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## Homework Statement

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

## Homework Equations

KEi + PEi = KEf + PEf (Initial kinetic energy + Initial potential energy of gravity = Final kinetic energy + final potential energy of gravity) rewritten as:

1/2mvi^2 + mgyi = 1/2 mvf^2 + mgyf

where:

vi = initial velocity

m = mass

g = acceleration of gravity

yi = initial y-position

vf = final velocity

yf = maximum height/final y-position

## The Attempt at a Solution

I think I have this solved correctly, and I was wondering if anyone would be willing to confirm the answer I got.

These are the values I substitute:

vi = 40 m/s

g = 9.8 m/s2

yi = 0 m

vf = 40 cos(60) (my rational behind this is that at maximum height, the velocity is only in the positive x-direction)

yf = unknown solving for.

First, I cancel out the mass in the equation by dividing the entire equation by mass.

Next, I substitute values, ending up with:

1/2(40^2) = 1/2 (40 cos60)^2 + (9.8)yf

800 = 200 + 9.8yf

600 = 9.8yf

yf = 61.2 meters