Uniform Acceleration/Projectile Motion

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The discussion focuses on a physics problem involving a long jumper's jump, requiring calculations of horizontal and vertical velocities, as well as the projection angle. The horizontal velocity at take-off is calculated to be 9.38 m/s based on the jump distance and time in the air. The participant struggles with determining the vertical velocity, needing to apply the equation for vertical motion, given the center of gravity's drop. Assistance is provided to clarify the use of the vertical displacement in calculations. The conversation emphasizes the importance of understanding projectile motion concepts to solve the problem effectively.
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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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