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

[bionut]

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??

 

[bp_psy]

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.

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.

 

[lewando]

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

 

[bionut]

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...

 

[bionut]

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??

 

[lewando]

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.

 

[lewando]

v=3.13

Would I be correct??

Yes, in m/s.

 

[bionut]

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??

 

[lewando]

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