# Another motion prob

1. Sep 13, 2007

### faoltaem

$$_{}$$1. The problem statement, all variables and given/known data

If a long jumper at the top of his projectory is moving at 6.5 m/s (horizontally) and his cantre of mass is 1.1m above where it was when he launched into the jump, how fast must he have been moving when he launched?

2. Relevant equations

v$$_{x}$$ = v$$_{0x}$$ + a$$_{x}$$t
x = $$\frac{1}{2}$$ (v$$_{0x}$$ + v$$_{x}$$)t
x = v$$_{0x}$$t + $$\frac{1}{2}$$a$$_{x}$$t$$^{2}$$
v$$_{x}$$$$^{2}$$ = v$$_{0x}$$$$^{2}$$ + 2a$$_{x}$$x

3. The attempt at a solution

v$$_{x}$$ = 6.5 m/s
y = 1.1m
y$$_{0}$$ = 0m
while the jumper is at the top of trajectory $$\rightarrow$$ v$$_{y}$$ = 0m/s

is it possible to work out this question with just those equations?
they all have either a time, acceleration of x component (i.e. distance travelled)
also is this considered a projectile motion problem as he was running before he jumped.

2. Sep 13, 2007

### learningphysics

HInt: what is the horizontal velocity when he jumped? what is the vertical velocity when he jumped? Find these 2 separately... then you can get the velocity with which he jumped.

3. Sep 13, 2007

### faoltaem

thanks learningphysics

cool so it's:

vy$$^{2}$$ = voy$$^{2}$$ - 2g$$\Delta$$y
0$$^{2}$$ = voy$$^{2}$$ - 2$$\times$$9.81$$\times$$1.1
voy$$^{2}$$ = 21.582
voy = 4.65 m/s

vx = vox
voy = 6.5 m/s

vo = $$\sqrt{voy$$^{2}$$ + vox$$^{2}$$}$$
= $$\sqrt{4.65$$^{2}$$ + 6.5$$^{2}$$}$$
= $$\sqrt{21.582 + 42.25}$$
= $$\sqrt{63.832}$$
= 7.989 m/s $$\rightarrow$$ 8 m/s

4. Sep 13, 2007

Looks good!