Finding the Upward Velocity of a Long Jumper's Flight Phase

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Homework Help Overview

The discussion revolves around determining the upward velocity of a long jumper during the flight phase of a jump, where the jumper rises 0.5 meters and has a forward velocity of 8 m/s. Participants are exploring the relationship between vertical and horizontal components of motion using trigonometric and kinematic principles.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the use of trigonometric relationships to find the upward velocity, questioning the assumptions made about angles and components of velocity. Some suggest using kinematic equations instead, while others express confusion about the relationship between the vertical and horizontal components.

Discussion Status

There is an ongoing exploration of different methods to approach the problem, with some participants offering guidance on isolating vertical motion from horizontal motion. Multiple interpretations of the problem are being considered, particularly regarding the need for angles in calculations.

Contextual Notes

Some participants note that the information provided may be insufficient to solve the problem directly, and there is discussion about the assumptions regarding the total velocity versus its components.

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Homework Statement



A long jumper rises 0.5 meters during the flight phase of his jump. His forward velocity is 8 ms. What is the upward velocity?

Homework Equations





The Attempt at a Solution



Vv=Vsin\vartheta

So \vartheta is the unkown

Okay, I did a skectch to work out basic trig, with 8m/s as my H and 0.5m as my 0
using sin\vartheta=o/h, =0.5/8 = 3

Then VV=8Sin3 = 0.38 m/s (which is wrong, but where did I go wrong??
 
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bionut said:
A long jumper rises 0.5 meters during the flight phase of his jump. His forward velocity is 8 ms. What is the upward velocity?

Is this the exact wording ? If it is then I don't think you have enough info to solve it.

bionut said:
Vv=Vsinϑ

So ϑ is the unkown

Okay, I did a skectch to work out basic trig, with 8m/s as my H and 0.5m as my 0
using sinϑ=o/h, =0.5/8 = 3

Then VV=8Sin3 = 0.38 m/s (which is wrong, but where did I go wrong??
This does not make sense. You try to find an angle between a velocity and distance when you need an angle between velocities. you also assume forward velocity to be the total velocity not just the vertical component.
 


What is preventing you from using one of the standard kinematic equations? Hint: you know vy, ay, y, and y0.
 


Hi, yea the exact words are "A long jumper rises 0.5 meters during the flight phase of his jump. His forward velocity is 8 ms. What is the upward velocity? " I assumed because its asking for upward velocity = Vertical component which is Vsin(theata) so Theta is the unknown?

The next questions after that asks "A long jumper rises 0.5 meters during the flight phase of his jump. His forward velocity is 8 ms. What is the angle of take off?"

So beacuse off this I assume I have the wrong idea regarding the first part; hence I need to not find the angle to calculate the upward velocity... if so I am confused...
 


oh thanks, so if I use 0=Vi^2 + 2ad

0=Vi^2 + (2 x 9.81 X .05)
Vi^2=9.81
v=3.13

Would I be correct??
 


Isolate the activity in the +y direction from the activity in the +x direction. Now ignore the activity in the +x direction. What's left is the jumper is jumping up 0.5m.
 


v=3.13

Would I be correct??

Yes, in m/s.
 


Hi again... the second part to the question "A long jumper rises 0.5 meters during the flight phase of his jump. His forward velocity is 8 ms. What is the angle of take off?"

Okay, I did a skectch to work out basic trig, with 8m/s as my H and 0.5m as my 0
using sinϑ=o/h, =0.5/8 = 3
but it should be 21.6 ... do you know where I went wrong??
 


bp_psy has pointed out that your vectors are not consistent. Since you now know Vy and Vx is given, work with them. If you want to use sinϑ=o/h, then you need to compute the magnitude of the Vx + Vy combined. Or you could just run with tanϑ=o/a.
 
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