Calculating Maximum Height of an Arrow with Loss of Mechanical Energy

[Lil Uzi Vert]

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?

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}

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!

 

[kuruman]

Assuming a 20% loss in Mechanical Energy

30% loss of mechanical energy, this is what i am thinking

Where does the 30% come from? Your method is correct.

 

[gneill]

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]

Did I do this question right or wrong?

Yes it's the right approach just check the working as others have suggested.

 

[Lil Uzi Vert]

Where does the 30% come from? Your method is correct.

i meant 30%

 

[Lil Uzi Vert]

Check that calculation. Did you divide by two?

(In the post i meant 30% loss, not 20 % that's 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]

(In the post i meant 30% loss, not 20 % that's 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