# Kinematics in Two Dimensions?

1. Sep 4, 2004

### pinky2468

Kinematics in Two Dimensions??

Hi, I was wondering if anyone could give a little help on this problem. I can't figure out where to begin!
An Olympic long jumper leaves the ground at an angle of 23 degrees and travels through the air for a horizontal distance of 8.7m before landing. What is the take off speed of the jumper?
Thanks!!

2. Sep 4, 2004

### Tide

Let x be horizontal and y be vertical and suppose the jumper takes off from the origin (0, 0) at t = 0. Then
$$x = v_0 \cos( \theta) t$$
and
$$y = v_0 \sin(\theta) t - \frac{1}{2}g t^2$$
so, first, find the time at which he/she lands (y = 0) and use that value of t in the first equation to find $v_0$

3. Sep 4, 2004

### pinky2468

Why is y=0 and for the y equation given is V=0?

4. Sep 4, 2004

### HallsofIvy

Staff Emeritus
y=0 on landing because he is landing on the ground which where you are measuring from!

No, v0 is the same in both equations. (It is the sine and cosine that distinguish vertical and horizontal speeds.) That means that what Tide suggested isn't quite right- you can't "find the time at which he/she lands (y = 0) and use that value of t in the first equation to find v0".

What you can do is use the length of the jump given and solve the two simultaneous equations
$$v_0cos(23)t= 8.7$$ and
$$v_0sin(23)t- \frac{g}{2}t^2= 0$$
for both t and v0.

5. Sep 4, 2004

### pinky2468

I'm sorry, I feel retarded but I still don't understand how you can have to unknown variables in one problem. If I don't know Vo or time how do I even solve the first equation?

6. Sep 4, 2004

### needhelpperson

you can always remove one variable by substituting it as something else....

for instance, you can use t = x/(cos23 *v)

Last edited: Sep 4, 2004
7. Sep 4, 2004

### Tide

Picky, picky!

Obviously, when I say "solve for" I meant solve for t in terms of [itex]v_0[itex].