Uniform Acceleration/Projectile Motion

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Homework Help Overview

The problem involves a long jumper's performance, specifically analyzing his jump distance, time in the air, and the vertical drop of his center of gravity. The subject area includes concepts of uniform acceleration and projectile motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the horizontal velocity and vertical velocity at take-off, expressing uncertainty about how to proceed without knowing the projection angle. Some participants suggest using kinematic equations to find the vertical velocity.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the vertical velocity calculation. There is recognition of confusion regarding the center of gravity, and some clarification has been offered, but no consensus has been reached on the overall solution.

Contextual Notes

Participants are working within the constraints of the problem statement, which includes specific measurements and a focus on deriving velocities and angles without additional information about the projection angle itself.

barnaby
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Homework Statement


"A long jumper's jump measures 7.5m, and he is in the air for 0.8s. His centre of gravity falls 0.95m between his take off and landing."

a) Calculate his horizontal velocity at take-off.
b) Show that his vertical velocity is 2.7(3) m/s.
c) Hence calculate the angle at which he projects himself at take-off.

Homework Equations



v_{ave} = \Delta x / \Delta t
v_{horiz.} = v_{result.}cos \Theta
v_{vert.} = v_{result.}sin \Theta

There may be some more that have to come into use, too...

The Attempt at a Solution



I calculated that his horizontal velocity at take-off is 9.38m/s, since he travels 7.5m horizontally in 0.8s.

I have spent about an hour trying to work out what to calculate, and how, in order to be able to calculate the vertical velocity at take-off.

I tried to calculate his instantaneous velocity at the point where his centre of gravity had reached its original point, but I couldn't since I don't know the angle at which he was projected...

It all goes downhill from here, really.

Any help would be greatly appreciated.

Thanks a lot.
 
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d = v1*t + (1/2)at^2

apply this equation for the vertical direction. you have t, a and d. you can solve for v1.
 
Oh, of course... the centre of gravity thing threw me a bit. Thanks!
 
Last edited:
barnaby said:
Oh, of course... the centre of gravity thing threw me a bit. Thanks!

No prob. just use d = -0.95m
 

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