Solve for Speed at Start of Long Jump: 6.5 m/s

[faoltaem]

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Homework Statement

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?

Homework 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

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.

 

[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.

 

[faoltaem]

thanks learningphysics

cool so it's:

vy^{2} = voy^{2} - 2g\Deltay
0^{2} = voy^{2} - 2\times9.81\times1.1
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

 

[learningphysics]

Looks good!