physics
posted by rmj on .
I'm having trouble getting the answer for the 2nd and 3rd questions.
# A soccer ball is kicked from the ground. When it reaches a height of 5.93 m, the velocity is v=6.45i+7.46j m/s.
1. Determine the maximum height that the ball reaches. (the right answer is 8.77 m)
2. Determine the total horizontal distance travelled by the ball.
3. What is the speed of the soccer ball just before it hits the ground?
for number 2, i found the total time from yf = yi + viyt  1/2gt^2
and plugged it into vx=dx/t to get the horiz distance but i still can't
get the right answer
did i miss anything?

The time the ball is in air is
2*Vyo/g , where Vyo is the initial vertical velocity component. Solve for that first.
From energy considerations,
7.46^2/2 + gH = Voy^2/2
Voy^2/2 = 86.0 m^2/s^2
Voy = 13.11 m/s
so T = 2.67 s
The distance travelled in that time is
Vx*T = 17.24 m 
The way I solved the first part of the question was:
Vo=?
d=7.18 m
Vf = 7.80m/s
a = 9.81m/s^2
So find the initial Vo velocity using:
Vf^2  Vo^2 = 2ad
Then once you get Vo, use the same equation to find the d, when Vf = 0, that's when the height is maximal.
Hope that helps.
Cheers,
Ross 
drwls  could you please explain the reasoning behind this:
"From energy considerations,
7.46^2/2 + gH = Voy^2/2
Voy^2/2 = 86.0 m^2/s^2
Voy = 13.11 m/s
so T = 2.67 s
What does "H" represent? Why can't you just use the total time that the ball is in the air, and multiply that by the initial xcomponent of velocity? Since d=vt 
the G represents gravity, and H represents the height