# Athlete jumps at angle with distance find speed

1. Sep 26, 2010

### afa

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

Leaves the ground at a 33.6 degree angle and travels at 7.77 m. What is take off speed?
If speed were increased by 4% how much longer would the jump be?

2. Relevant equations

Vx=Vcos(theta)

3. The attempt at a solution

7.77=vcos33.6 got 9.33 but says it wrong, then I added 4% to this and did 9.7cos33.6 which also seems to be wrong..what am I missing?

2. Sep 26, 2010

### Juggernaut

Time my friend-time. What`s the time what it takes to travel.

3. Sep 26, 2010

### JaredJames

Hi afa, could you be a little more specific please, you say he travels 7.77m, but where? Horizontally / Vertically / Diagonally? Is this all of the question or is there more detail to it?

Jared

4. Oct 16, 2010

### dmkeddy

I'm having a similar problem:
An athlete executing a long jump leaves the ground at an angle of 30 degrees and travels 8.90m. What was the take-off speed?

5. Oct 16, 2010

### JaredJames

Both responses from juggernaut and myself above apply to this second problem as well.

Jared

6. Oct 16, 2010

### dmkeddy

travels horizontally 8.90m, no vertical displacement, time is unknown, how do I solve for two variables? Or what do I substitute velocity or time with to solve for the other?

7. Oct 16, 2010

### JaredJames

Leaves the ground at 30 degrees, but there's no vertical displacement?

8. Oct 16, 2010

### JaredJames

You really do need the time factor, and based on what you have given you can't calculate it as far as I can tell.

I'm not sure if you can solve with only two variables. Try rearranging your equations of motion to get something you can solve (perhaps simultaneously).

You can't just substitute time or velocity as they'll generate different answers.

dmkeddy, you should start your own threads for things like this and not hijack an old one as it will gain you a better response.

Last edited: Oct 16, 2010
9. Oct 16, 2010

### lvslugger36

Since you have to assume that his y displacement is zero by the time he lands, you could use the range equation for this which is (v^2)sin(2x)/g. You know the range is 7.77 and you know x, which is the angle. So solve for v. For the second case, all is being kept constant, except for velocity. So the equation should be kept normal but except for where you write v^2 write (1.04v)^2. If you simplify, you'll note that everything is the same except for the 1.04^2. Hence, just multiply 7.77 by 1.04^2.