# Kinematics in 2-dimesions

1. Jan 30, 2008

### ian_durrant

[SOLVED] Kinematics in 2-dimesions

1. The problem statement, all variables and given/known data

A soccer player kicks the ball toward a goal that is 25.0 m in front of him. The ball leaves his foot at a speed of 19.2 m/s and an angle of 65.8 ° above the ground. Find the speed of the ball when the goalie catches it in front of the net. (Note: The answer is not 19.2 m/s.)

2. Relevant equations

x=V(0)t+(1/2)gt^2

V^2=V(0)^2+2gy

V= sqrt(Vx^2+Vy^2)

3. The attempt at a solution

Ok let me define all the variables I've got so far

V(0)= 19.2

Vx0= 7.87
x= 25 (12.5 for one half)
Vx= ?(at top of apex )

Vy= 0 (at top of apex)
gy=-9.8
y= 15.626
Vy0= 17.5 (initial y veolicty)

Ok so the first thing I did was find the inital velocity for the x and y componets using sine and cosine, so

V0x= 19.2cos65.8= 7.87
V0y=19.2sin65.8= 17.5

I then used my inital y velocity to find out the heigth (y distance) of the shot

Vy^2= Vy0^2+2gy
0= (17.5)^2+2(-9.8)y
y= 15.625

I'm unsure where to go now in order to out the speed of the ball when the goalie catches it i know that V= sqrt (Vx^2+Vy^2). Should i try and figure out the two final velocities and then plug them into that equation?

Last edited: Jan 30, 2008
2. Jan 30, 2008

### blindside

You need to find the time at which the ball reaches the position of the net. What you've written as $$g_x$$ is actually just acceleration in the x-direction, g is simply used to denote gravitational acceleration (which only occurs in whats defined as the y-direction in this problem). If air resistance is ignored, acceleration in the x-direction is 0.

Thus you know speed in the x-direction will not change over the course of the motion. Once you've found the time, you need to find the x and y velocities, and as you said, the square root of the sum of their squares is the speed at that point.

3. Jan 30, 2008

### ian_durrant

Ahhh i see it now, the height wasn't even needed in this problem, thanks for the tip.