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Arrow Loses Mechanical Energy

1. Homework Statement
A 125 - G arrow is shot vertically upwards with a Vi of 28/ms. Assuming a 20% loss in Mechanical Energy while ascending, what maximum height above the position it was shot does it reach?

2. Homework Equations
Vi= 28m/s
Vf= O m/s
D= ?
a= -9.8 m/s

Vf^2= Vi^2 + 2aD
Eg = mgh
Ek = 1/2mv^2
EM = Eg + Ek (not sure}
3. The Attempt at a Solution
I did not really understand this question but I tried to make a solution using what I know.
I tried using one of the 5 equations but that did not help so i tried using work and energy equations.

So i did Ek first
Ek= 1/2 mv ^2
Ek = 1/2 (0.125 kg) (28 m/s)^2
Ek = 98 J

Next I tried finding the Eg
Eg = Ek x 0.70
(30% loss of mechanical energy, this is what i am thinking)
Eg = 68.6 J

Eg = mgh
Eg= 0.125 (9.8) h
68.6 = 1.225 h
68.6/1.225= h
h= 55.51 m

Did I do this question right or wrong? I am very confused. Some help would be appreciated!
 

gneill

Mentor
20,486
2,610
So i did Ek first
Ek= 1/2 mv ^2
Ek = 1/2 (0.125 kg) (28 m/s)^2
Ek = 98 J
Check that calculation. Did you divide by two?
 

CWatters

Science Advisor
Homework Helper
Gold Member
10,528
2,291
Last edited:
Check that calculation. Did you divide by two?
(In the post i meant 30% loss, not 20 % thats why i did multiplying by 0.7.)
I divided by two so the number is 49 J , multiplying it by 0.7 would be 34.3 J
34.3 J / 1.225 = h
H= 28 M exactly
Is this method now correct?
 

gneill

Mentor
20,486
2,610
(In the post i meant 30% loss, not 20 % thats why i did multiplying by 0.7.)
I divided by two so the number is 49 J , multiplying it by 0.7 would be 34.3 J
34.3 J / 1.225 = h
H= 28 M exactly
Is this method now correct?
Yes. Although "exactly" depends upon your choice of value for g :smile:
 

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