What Was the Take-Off Speed of the Long Jumper?

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Homework Help Overview

The problem involves determining the take-off speed of an athlete executing a long jump, where the jump is characterized by an initial angle of 30 degrees and a horizontal distance of 8.90 meters. The context is rooted in kinematics, particularly projectile motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between vertical and horizontal components of motion, questioning the assumptions about vertical displacement and the need for vertical height in the calculations. There are attempts to clarify the equations of motion relevant to both vertical and horizontal trajectories.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem setup. Some guidance has been offered regarding the separation of vertical and horizontal motion, but there is no explicit consensus on how to proceed with the calculations.

Contextual Notes

There is a noted confusion regarding the vertical displacement being zero, with participants emphasizing that there must be a vertical height involved in the jump. The lack of flight time information is also highlighted as a constraint in solving the problem.

dmkeddy
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Homework Statement



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?

Homework Equations



d(vertical)=(vi)(t) + 0.5at^2

The Attempt at a Solution



0=Vsin30t + (0.5)(-9.8)(t^2)

I cannot solve for two separate variables, how can I solve this problem?
 
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EDIT: I see what you've done.

Right, in your attempt, d does not equal 0. There has to be some vertical height.

For horizontal you know:

d = 8.9m

For vertical you know:

a = -9.8 for the ascent and 9.8 for the descent.

You know vf for the ascent = 0 and vi for the descent = 0.
 
Last edited:
that is the vertical, there is no horizontal acceleration
Vertical:
Vi=?
Vf=?
a=-9.8m/s^2
t=?
d=0

Horizontal:
Vave=?
t=?
d=8.90m
 
dmkeddy said:
that is the vertical, there is no horizontal acceleration
Vertical:

d=0

d does not equal 0. There has to be vertical height.

The vertical phase is split in two. See previous post for details.

Unless you know the flight time, this question all comes down to the vertical components.
 

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