# Soccer Ball Projectile Prblm

1. Feb 3, 2009

### afg_91320

1. The problem statement, all variables and given/known data
A soccer ball is kicked at an angle of 15 degrees with the ball traveling at 15.0 m/s
a) what is the max height reached by the ball?
b) max range?
c) how could range be increased?

so i need to find:
ymax
R = xmax

2. Relevant equations
these i assume are the given:
$$\vartheta$$=15 degrees
v0=15.0 m/s
ay= -g
x0, y0 and y = 0

3. The attempt at a solution
vx0 = v0 cos 15 = 15 cos 15 = 14.5
vy0 = v0 sin 15 = 15 sin 15 = 3.88

vy = 0 = vy - gtu <---tu is time of ball movin upward

must get tu:

= vy0 / (g)

here is where i get stuck cuz i dont know what to do with the g since i dont have a value for it.

2. Feb 3, 2009

### needlottahelp

g is acceleration due to gravity. It is about 9.8 m/s^2

3. Feb 3, 2009

### afg_91320

so g is always 9.8?
is that just a given?

what do i do next?

4. Feb 3, 2009

### needlottahelp

yeah. i think sometimes g is rounded to 10 or 9.81. i think so unless the problem states otherwise. So use that to find the time.

If you want the max height, you should probably use this kinematic equation:

vyf^2 = vyi^2 + 2at

Note: a = -g in this case

For the range, you use d = (vx) (t) because there is no acceleration in the horizontal

5. Feb 3, 2009

### tnutty

max range occurs at angle 45 degrees

6. Feb 4, 2009

### afg_91320

so i got for the max height: 0.772 m <--thats about 3 ft. sounds right for a soccer ball kicked at 15 degrees?

and max range to be: 11.5 m