How Can Range be Increased for a 15-Degree Soccer Ball Projectile?

[afg_91320]

Homework Statement

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:
y_max
R = x_max

Homework Equations

these i assume are the given:
\vartheta=15 degrees
v₀=15.0 m/s
a_y= -g
x₀, y₀ and y = 0

The Attempt at a Solution

vx₀ = v₀ cos 15 = 15 cos 15 = 14.5
vy₀ = v₀ sin 15 = 15 sin 15 = 3.88

v_y = 0 = v_y - gt_u <---t_u is time of ball movin upward

must get t_u:

= v_y0 / (g)

here is where i get stuck because i don't know what to do with the g since i don't have a value for it.

 

[needlottahelp]

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

 

[afg_91320]

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

what do i do next?

 

[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

 

[tnutty]

max range occurs at angle 45 degrees

 

[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