A ball is kicked from the ground at an angle of 65° and at a speed of 8 m/s. What is its speed when it reaches 1/2 of the maximum height along the path, assuming that air resistance is negligible?
My head hurts from this.
I made a diagram of this event by using a vector pointing diagonally with 65 degrees to the horizontal
I used trig to find the velocity at the x and y components
sin65 = y/8
Y = 7.25m/s
x = 3.38m/s
from here, i didn't know what to do. I used the equation V^2 = Vi^2+2a[tex]\Delta[/tex]x
[tex]\Delta[/tex]x = .58287751m
Then, to obtain the height
H = 1.25m
then half of that is .6249m
and uh....conservation of energy
Intial 1/2mVi^2 = Final mgh + 1/2mVf^2
You can cancel out the m's
1/2(8)^2 = (9.80)(.6249m) + .2Vf^2
Vf= 7.19 = 7.2m/s ?