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

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SUMMARY

The discussion centers on calculating the upward velocity of a long jumper who rises 0.5 meters during his flight phase with a forward velocity of 8 m/s. The correct upward velocity is determined using kinematic equations, specifically 0 = Vi² + 2ad, leading to an upward velocity of approximately 3.13 m/s. Participants clarify that the angle of takeoff is not necessary for calculating the upward velocity, and emphasize the importance of isolating vertical and horizontal components of motion. The correct approach involves using known values of vertical displacement and acceleration due to gravity.

PREREQUISITES
  • Understanding of kinematic equations, specifically 0 = Vi² + 2ad
  • Basic knowledge of trigonometry, particularly sine and tangent functions
  • Familiarity with vector components in physics
  • Concept of vertical and horizontal motion separation
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  • Study the application of kinematic equations in projectile motion
  • Learn how to resolve vectors into their components
  • Explore the relationship between vertical and horizontal velocities in projectile motion
  • Investigate the calculation of angles of takeoff using trigonometric functions
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Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators seeking to clarify concepts related to velocity components and kinematic equations.

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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.
 
Last edited:

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