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Verticle Jump 
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#1
Dec1208, 08:41 PM

P: 32

1. The problem statement, all variables and given/known data
In order for a volleyball player to jump vertically upward a distance of 0.8 meters, his initial velocity must be? 2. Relevant equations s=v_{0}t + .5at^{2} vf= v_{0} + at 3. The attempt at a solution I missed the question the other day on a practice and it has been forever since I messed with this material. Those equations were given as well as the answer, the only problem is I do not know what the letters represent or even how to solve for anything with only information of 0.8 meters. Could someone help me learn this and what the expressions mean so I can solve it? 


#2
Dec1208, 08:46 PM

P: 145

Keep in mind the direction of g in reference to the player's direction of motion, i.e. positive or negative?
Also, do you have another variable besides v0 that you need but don't have? Try to solve it in terms of another variable, which kills two birds with one stone. EDIT: Oh, and the variables mean s, the displacement v0, the initial velocity t, the change in time a, the acceleration of the body (g = 9.8 m/s^2) vf, the final velocity Always find out what the key variables mean in a lesson before solving the problems. 


#3
Dec1208, 08:51 PM

P: 32

ooo thank you so much for laying out what the variables mean...let me see what I can do now...
Let me try and have a crack at this again. 


#4
Dec1208, 08:53 PM

P: 145

Verticle Jump
Displacement is the same as distance in this case. Your distance, vertical, is given to be 0.8 m.
g IS your acceleration a. EDIT: Oh, and I saw your other thread. You must be lacking a few concepts without an instructor or textbook... When a body reaches a maximum height (i.e., it falls down instantly afterwards) in vertical motion, the velocity (final) at that instant is equal to zero. 


#5
Dec1208, 09:13 PM

P: 32

Ya...I have that down...but I am having a problem...lets see...
v_{f}= v_{0} + at so 0 = v_{0}+9.8t or am I doing gravity wrong? Cause I solved for v_{0} and moved over to the other equation... 0.8 = 9.8t^{2} + .5(9.8)(t^{2}) .8 = 9.8t^{2} + 4.9t^{2} (Now if I already have not made an error with a before this is where I am getting stuck) .8 = 4.9t^{2} I cannot remember how to get t alone from this point (bad I know) 


#6
Dec1208, 09:20 PM

P: 145




#7
Dec1208, 09:32 PM

P: 32

Ok so it is
0.8 = 9.8t^{2}+(.5)(9.8)(t^{2} 0.8 = 4.9t^{2} .163 = t^{2} .4 = t 0 = v_{0}  9.8(.4) 0 = v _{0}  3.92 3.92 = v_{0} Now all I need is more than this one problem to practice lol...you are awesome, thank you, thank you, thank you. 


#8
Dec1208, 09:37 PM

P: 145

Students helping students :D
By the way, you could have done the problem in 2 steps instead of 3 steps if you had solved for t instead of solving for v0 first. Generally with algebra, to get rid of a variable you solve for that variable. 


#9
Dec1208, 09:42 PM

P: 32

A) Now I know what these variables mean... B) Memorized some formulas lol I will try making up some problems later and solve in 2 steps... It has been sooo long since I messed with math or anything. 


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