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Homework Help: Projectile Motion/ Air Resistance Question

  1. May 19, 2015 #1
    I actually have 2 questions that I am getting stuck on around the same point..

    Question 1) An arrow with a mass of 80g is fired at an angle of 30 degrees to the horizontal. It strikes a target located 5 m above the firing point and impacts the target traveling at a speed of 20 meters/sec. If 10% of the initial energy of the arrow is lost to air resistance, what was the initial speed of the arrow?

    My attempt at the problem:

    Energy final= Kf + Uf = 1/2*m*vf^2 + m*g*hf

    Energy initial= Ki + Ui = 1/2*m*vi^2 + m*g*hi (but since initial height is 0)= 1/2*m*vi^2

    Wnonconservative= Ei- Ef= 1/2*m*vi^2 -(1/2*m*vf^2 + m*g*hf)

    And since 10% of the initial energy is lost,
    Wnonconservative= 0.1* Ei= 0.1* (1/2*m*vi^2)

    This next step is where I am getting confused, in solving for the initial speed:

    Wnonconservative= 1/2*m*vi^2 -(1/2*m*vf^2 + m*g*hf)


    0.1* (1/2*m*vi^2)= 1/2*m*vi^2 -(1/2*m*vf^2 + m*g*hf)

    The solution in the book is 23.5 m/s, but I don't know how to get it from here. Please help!

    Question 2) Two different objects are dropped from rest off of a 50 m tall cliff. One lands going 30% faster than the other, and the two objects have the same mass. How much more kinetic energy does one object have at the landing than the other?

    My attempt at the problem:


    Vfinal, faster object= 1.3

    Vfinal, slower object=1

    initial height=50 m

    final height= 0


    So the difference in kinetic energy is:

    1/2 *m *(1.3v)^2 - 1/2 *m *v^2= ?

    The answer is 69% more kinetic energy for the faster rock, but I'm having trouble finding out how to get that from here. Any help is very much appreciated! Thanks!
  2. jcsd
  3. May 19, 2015 #2


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    You have all of the values except vi. Simply solve for vi and insert the values in the resulting equation.

    You are looking for the ratio of the kinetic energies, not their difference.
  4. May 19, 2015 #3
    Thank you! I got the first question, but for the second question, I'm not sure if I'm using the correct method to get the answer..

    1/2* m * (1.3v)^2 - 1/2* m * v^2 = (supposed to equal) 0.69 * 1/2 * m * v^2, but how do I get that....

    1/2* m * 1.69* v^2 - 1/2* m* v^2
    = 0.845* m*v^2 - 0.5 * m* v^2
    = 0.345 m*v^2 .... which if you multiply by 2 you get 0.69, so
    = 0.69 * 1/2 * m * v^2

    There we go! I think I got it now thank you :)

    Is there any easier way to solve then doing all of that? Because it'd be tricky to remember to have to multiply the 0.345 * 2 to get 0.69 * 1/2 * m * v^2....but I guess I'll just have to remember to solve to get 1/2 * m * v^2 since we are solving for kinetic energy
    Last edited: May 19, 2015
  5. May 19, 2015 #4


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    Yes, taking the ratio of the kinetic energies. Everything apart from ##1.3^2 = 1.69## will cancel.
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