Range Question

  • #1
I am having the hardest time getting my range to match (at least I think they should). I used two range equations for the same problem as demonstrated below. I did it both ways to check my answer, and they are different. I was hoping you could tell me where I went wrong...I think my time of flight is incorrect.

v_0=105m/s
h=125m
Θ=37degrees
v_oy=63.19m/s (105m/s sin37)
v_ox=83.86m/s (105m/s cos37)

For my time of flight I used the equation y = v_0yt - 1/2gt^2 - h

Plugging in the numbers I get y = 63.19t - 1/2(9.8m/s)t^2 - 125

Rearranging the equation for quadratic, -4.9t^2 + 63.19t - 125

Then I use my quadratic formula and get
(-63.19(+-)39.28)/-9.8 thereby getting +2.349s and +10.45s. Neither of these fit when multiplied by my x of 83.86m/s. I get 196m and 876m.

For the problem I am checking the answer with is derived from the equation
R = v_o^2(sin2Θ)/g
So I get (105m/s)^2(sin2(37))/9.8m/s = 1081.41m

They don't match and they should, right?
 

Answers and Replies

  • #2
HallsofIvy
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I have a question. Are you firing this thing from a hole in the ground? You tell us that "h= 125m" which, I thought, meant that the projectile started from 125m above the ground (y= 0). But you use the formula y = 63.19t - 1/2(9.8m/s)t^2 - 125, which, for t= 0, gives y= -125. Shouldn't your formula be y = v_0yt - 1/2gt^2 + h ? (That's why you are getting TWO positive values for t when y= 0! Once on the way UP and the other on the way DOWN. Since the projectile starts above y=0, you should only get to y=0 on the way down.)
 
  • #3
Not firing from a hole; it's from the top of a building. Thank you for pointing out my mistake in signs. I fixed it and got 14.63seconds for my time of flight. However, that still does not match the R of 1,081.46m. When I multiply my time by v_0x I get 1226.87m. Why won't they match? What else am I doing wrong because I just don't see it.
 
  • #4
HallsofIvy
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Did you notice that there is no "h" (initial height) in the
R = v_o^2(sin2È)/g formula? That gives the correct range only when firing from "ground level" (i.e. initial height and final height are both 0) (more generally, both the same).

If you are firing from ground level (initial point is (0,0), then x and y are given by
x= v_x t and y= v_y t- (g/2)t^2= t(v_y- (g/2)t)

It is easy to see that y=0 at t=0 and at t= 2v_y/g.
(The more general case, initial y NOT 0, is harder!)

The x= v_x(2v_y/g)= 2(v0 cos(theta))(v0 sin(theta))/g
= v0^2 (2cos(theta)sin(theta))/g
and of course you know that 2cos(theta)sin(theta)= sin(2theta)!

x=
 
  • #5
I realize now that you cannot use range formula R=V_o^2(sin2θ)/g when firing from the top of a building since it is not of perfect parabolic shape. I should only use the equation R=V_ox(t) to find the range.
 
  • #6
HallsofIvy
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No, the trajectory (because you are ignoring air resistance) IS a "perfect parabola". Because the initial point is not at the same height as the final point, you do not have the symmetry you would have if that were true. Is that what you meant?
 

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