Vertical Jump Test and physics Check My Work

[Winner]

Ok :rofl: ,
I had to test my maximum vertical jump. I did three trials and here's what I got:
-Maximum standing height (ie without jumping reaching with one hand) = 221.5 cm
-Maximum jumping height with one hand (best of three) = 270 cm
-Net Height 270-221.5 cm = 48.5 cm, 0.485m
My weight: 68 kg
And now the physics I had to do:

1)Calculate maximum potential gravitational energy during jump.
So PE=mgh
=(68kg)(9.81)(0.485m)
=323.5 J

2)Calculate work done
W=F x d
= 68kg(9.81)(0.485m)
=323.5 J

3) Calculate the average force required to jump this high if you apply the force over a distance of 0.5 meters. <<<I don't get what this is trying to ask>>>

4) Calculate power produced to jump this high if force applied for 0.5secs.
P=work/t
=323.5j/0.5 s
=647 Watts

5)Calculate your vertical velocity half way back to the ground.
So that distance is is just half of 48.5 cm?? or 24.25 cm
323.5 J=mgh +1/2mv^2
323.5J= 68(9.81)(0.2425m) +1/2(68kg)v^2
so, 161.7(2)/68kg=v^2
4.76=v^2
**v=2.18 m/s??

6)Calculate your vertical velocity the instant before you land.
So the instand I land is when the h=0.01m or the smallest distance before hitting the ground.
323.5=68(9.81)(0.01m)+1/2(68)v^2
317(2)/68=v^2
9.32=v^2
**v=3.05 m/s?

7) Calculate the average force required to absorb the energy of your landing if you take 0.25 meters to stop.
<<<I also don't get what this one>>>
Alright thanks guys. :tongue:

 

[daniel_i_l]

It looks ok.
In question 6 I think that you can put in 0 for the hight because the want the speed the instant that you hit the ground.
In question 7 you can calculate the average force with the impulse (J = avgF/t). You can get the impulse from the change of momentum. In the end the speed is 0, that's why they asked for the final speed in question 6.

 

[Winner]

I'm surprised I got most of them lol. You jogged my memory about momentum though. Thanks. Any thoughts on question 3?

 

[Winner]

Wait, how am I suppose to find t? I know impulse is = Ft=m(v2-v1). I don't have t though to solve for F. I have everything else. :/

 

[daniel_i_l]

The first time I read it I assumed that it meant 0.25 seconds to stop. If it really means that you start stopping whenyour 0.25 meters in the air (though that dosen't really make much sense) then calculate the speed at that position and the time that it takes to fall from there.

 

[Winner]

I know, it doesn't make any sense, but that's exactly what it says on the worksheet.

I also don't get the wording on this one..

Calculate the average force required to jump this high if you apply the force over a distance of 0.5 meters. ?
Thanks.

 

[Winner]

I really don't get what this 0.25 meters come into play...I asked my teacher and he keeps telling me, use what info you got and find what you don't know, that's the key. LOL, uh yeah that's great.

 

[Winner]

Ok, I think I got it?


on the question where it asks:

Calculate the average force required to absorb the energy of your landing if you take 0.25 meters to stop.

Could it really be this simple?

I know my energy for the jump is 323.5J. And to apply such a force to stop at 0.25m, I'd just use the w=fd formula

and so f=323.5J/0.25m, f=1294N?
//////

Same to this question:

Calculate the average force required to jump this high if you apply the force over a distance of 0.5 meters.

W=fd
f=323.5J/0.5m
f=647N.

OR am I totally wrong and it has to do something with Ft=m(v2-v1)?

 

[Winner]

The other case...

Calculate the average force required to absorb the energy of your landing if you take 0.25 meters to stop.


F=m(v2-v1)/t
=68(0-3.05m/s)/t
=68(3.05m/s)/0.082
=2529N?

t=d/v
t=0.25/3.05
t=0.082

 

[Winner]

Anybody? it's due tomorrow, really need some advice. Thanks.