Calculating Takeoff Speed for an Olympic Long Jumper

[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!

 

[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 $v_0$

 

[pinky2468]

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

 

[HallsofIvy]

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

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.

 

[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?

 

[needhelpperson]

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

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

 

[Tide]

That means that what Tide suggested isn't quite right

Picky, picky!

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