Basic Kinetics

  • Thread starter Peach
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  • #1
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Homework Statement


At what height h above the ground does the projectile have a speed of 0.5v?


Homework Equations


K_1 + U_1 = K_2 + U_2
K = 1/2mv^2
U = mgh

The Attempt at a Solution


I made the point where the speed is .5v = point 1, and the maximum height = point 2.

Point 1:
K_1 = 1/2m(v_1x^2 + v_1y^2)
U_1 = mgh

Point 2:
K_2 = 1/2m(v_2x^2 + v_2y^2)
U_2 = mg(v^2/2g) = (mv^2)/2 <--maximum height, I got this from the previous part.

Then it's just plug and chug into the eqn K_1 + U_1 = K_2 + U_2 but I came out with the wrong answer which is (-v^2/8g). Oh I took out the v_1x^2 and the v_2x^2 because the x component remains constant throughout and I also took v_2y^2 because it's 0 at maximum height.


I must have done something wrong somewhere but I have no idea (probably my setup) so pls help?
 

Answers and Replies

  • #2
jamesrc
Science Advisor
Gold Member
476
1
Is that the whole problem statement? If it is, I'm assuming that this projectile is launched vertically from the ground with speed v.

In using the conservation of energy, find the total energy at the time of launch and at the time of interest. That is, call point "1" the time where speed = v and height = 0.

Then make point "2" the time where height = h (which is what you are solving for) and speed = .5v

I hope that helps. Let us know if you get stuck again.
 
  • #3
80
0
Thanks. I did that at first but then got stuck somehow, so I moved to maximum height. >< But I got it now. Thanks again.
 

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