Posted by Lindsay on .
The soccer goal is 23.05 m in front of a soccer player. She kicks the ball giving it a speed of 17.97 m/s at an angle of 25.83 degrees from the horizontal. If the goalie is standing exactly in front of the net, find the speed of the ball just as it reaches the goalie.
Do I need to use delta x = Vx x t to solve for time first? Then what?
I don't know the point of this problem: should air friction be ignored or not?
I think I would ignore air friction, which in reality is silly, but anyway.
Using the intial velocity and angle, solve for the initial vertical and horizontal velocities.
Using the vertical velocity, one needs to find the time of flight.
yf=yi+vyi*t - 4.9 t^2
4.9t^2-7.83+1=0 using the quadratic formula...
t= (7.83 +- sqrt (61.3-20)/9.8=1/45 sec
Now, where is the ball at that time?
d= 17.97cos25.83 * time=23.45 meters, within the goalie reach of .4 meters.
Now, what is the difference in energy?
Intial energy = 1/2 mv^2
final energy= 1/2 mvf^2 + mgh where h is one meter.
vf^2= vi^2 - 2gh
vf= sqrt (17.97^2 - 2*g*1)=17.42m/s
This problem is subject to a lot of differing assumptions, the main one is friction due to air. This assumes the ball is caught by moving the arms of the goalie.