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Hang timeby killerinstinct
Tags: time 
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#1
Sep1107, 07:39 PM

P: 88

An athlete jumps vertically. treat athlete as particle and Ymax is the maximum height above the floor the athlete achieves. To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2 to the time it takes him to go from the floor to that height.
we get 2 equations, Vymax/2 = V0Ygt_1 Vymax = Vymax/2  gt_2 after isolating t we get: 2t_2 / t_1 = 2 Vymax/2 / V0yVymax/2 V0Y  Vymax/2 < Vymax/2 <<<< how do you go about explaining this in common language?? the time it takes for person to reach ymax/2 from Yground is les thatn time it takes the person to reach Ymax from Ymax/2, so Vavg from Yground to Ymax/2 is greater thatn Ymax/2 to Ymax. 


#2
Sep1107, 08:51 PM

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P: 4,125

You already know that Vymax = 0
Use gravitational potential energy to calculate the velocity at ymax/2 in terms of v0 Then you'll know the velocity at the bottom... v0, velocity at ymax/2, and the velocity at ymax which equals 0. so then you can calculate the times... why did you 2t_2 / t_1 instead of t_2/t_1? 


#3
Sep1107, 08:55 PM

P: 88

i set t_2 as time from ymax/2 to ymax. so the time above ymax/2 is 2t2.
gravitational potential?? mgh = mv^2/2?? 


#4
Sep1107, 08:56 PM

P: 88

Hang time
on this problem, i'm only allowed to use kinematics. as much as i want to use energy, i can't.



#5
Sep1107, 09:12 PM

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P: 4,125

Yeah, mgh = mv^2/2 was what I was thinking. You can also get the velocity at ymax/2 using... vf^2 = vi^2 + 2as 


#6
Sep1107, 09:18 PM

P: 88

ya but t1 is time from bottom to middle. the question asks for 2t2/t1 not t2/t1. but anyways, thats technical... no big deal.
wouldn't using vf^2 = vi^2 +2as make this more complicated than it is?? i already got a ratio of 2Vymax/2 to VoyVymax/2 there is a way to simplify that?? by substituting in another equation. I dont see how it would simplify it. maybe i'm just bad at algebra, can you show me? 


#7
Sep1107, 09:32 PM

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P: 4,125

This doesn't simplify unless you write Vymax/2 in terms of V0y.... the question wants you to get the actual number for this ratio... Also be careful not to confuse Vymax/2 with (Vymax)/2 = 0 (since Vymax = 0)... might be better to write Vx where x = ymax/2... or something like that. 


#8
Sep1107, 09:38 PM

P: 88

vx^2=v0y^2 + 2 a ymax/2
you get: 2 sqrt (Voy^22gYmax/2) to Voy sqrt (V0y^22gymax/2) RATIO how do you simplify that?? 


#9
Sep1107, 09:42 PM

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#10
Sep1107, 09:55 PM

P: 88

Ymax = Voy^2/2g.
sub in and i get 2sqrt(2gYmax/2) to V0ysqrt(2gYmax/2) sub in V0y and i get 2sqrt(2gYmax/2) to sqrt (2gYmax)  sqrt (2gYmax/2) some algebra help... how you simplify whats in bold... if i'm doing it right.... i get 2 to 1 ratio?? that doesn't sound right. 


#11
Sep1107, 09:59 PM

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P: 4,125

You should get: [tex]\frac{2}{\sqrt{2}  1}[/tex] 


#12
Sep1107, 10:01 PM

P: 88

wait... i did it again and got ratio of
2 to sqrt2  1 is that right?? 


#13
Sep1107, 10:04 PM

P: 88

i guess i did it right..
i knew the physics of this problem. but i couldn't figure out the algebra. greatly appreciate your algebra lesson. thank you very much. 


#14
Sep1107, 10:25 PM

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#15
Sep1107, 10:25 PM

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