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

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The discussion revolves around calculating the upward velocity of a long jumper who rises 0.5 meters with a forward velocity of 8 m/s. Initial attempts to find the upward velocity using trigonometric functions were incorrect, as they confused the relationship between angles and velocity components. Participants suggested using kinematic equations to isolate vertical motion from horizontal motion, leading to the correct upward velocity of approximately 3.13 m/s. Additionally, the angle of takeoff was discussed, with guidance to use the correct vector relationships for accurate calculations. The conversation emphasizes the importance of separating vertical and horizontal components in physics problems.
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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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