Olympic Jumper

1. Feb 2, 2009

1. The problem statement, all variables and given/known data
An Olympic long jumper leaves the ground at an angle of 21.8 ° and travels through the air for a horizontal distance of 8.97 m before landing. What is the takeoff speed of the jumper?

2. Relevant equations
Square root of V^2_ox + V^2_oy

3. The attempt at a solution

Sqrt of 8.97^2 + (-9.8)2, idk what to do with the 28degree

2. Feb 2, 2009

LowlyPion

Your equation to combine the components of velocity doesn't exactly apply to the data given.

The problem gives you the distance of the jump. Not the horizontal component of V.

3. Feb 2, 2009

Delphi51

This is a projectile motion problem. You have to separate the vertical and horizontal motions because different formulas apply. The horizontal part is motion at constant speed while the vertical part is accelerated.

Use Vx = v*(21.8), Vy = v*sin(21.8)

For the horizontal part, use x = Vx*t
For the vertical part, use V = Vy + at and y = Vy*t + .5at^2
Put the numbers you know in all three formulas. It should then be possible to solve one of them and find something out (usually the time) so you substitute in another of them and find what you want.

4. Feb 2, 2009

wizard85

Let $$O(X,Y)$$ be a frame of reference with origin in the jumper. The equations of velocity and space in function of time are:

Velocity:

$$v_x=v_o cos\theta$$

and

$$v_y=v_o sin\theta + g*t$$

Space

$$s_x=v_o cos\theta * t +s_{ox}$$

and

$$s_y=\frac{1}{2}g*t^2 + v_o * sin \theta + s_{oy}$$

I assume $$s_{ox}$$ and $$s_{oy}$$ null inasmuch as origin is in the jumper. So the relation $$s_y=f(s_x)$$ by deleting $$t$$ is:

$$s_y=\frac{1}{2}g*\frac{s_x^2}{(v_o * sin \theta)^2} + s_x * tan\theta}$$

By substituting $$\theta=21.8°$$, $$s_x=8.97m$$and $$s_y=0$$ you'll get $$v_o$$

I hope this help, bye...

Last edited: Feb 3, 2009