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.
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
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