[SOLVED] Kinematics in 2-dimesions

ian_durrant

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)
gx= 0? (not so sure about this one)
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?

blindside

You need to find the time at which the ball reaches the position of the net. What you've written as [tex]g_x[/tex] 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.

ian_durrant

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