Answer check-derive a general formula for horizontal displacement

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Homework Statement



Derive a general formula for the horizontal distance covered by a projectile launched horizontally at speed vi from height, h.


The Attempt at a Solution



vyf = vyi -gt
t = vyf/g
x = vxi .t = vi cosΘ (t = vyf/g) = [vi^2 sinΘcosΘ]/g
 

Answers and Replies

  • #2
BvU
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The launch is horizontal, so your theta is known. You run into trouble with sin theta, which is 0.

Also, I don't find any dependence on h. Don't you think that if you launch from a high fortress, the projectile covers a bigger distance than if you launch from a small hill ?

Tip: make a drawing!
 
  • #3
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The launch is horizontal, so your theta is known. You run into trouble with sin theta, which is 0.

Also, I don't find any dependence on h. Don't you think that if you launch from a high fortress, the projectile covers a bigger distance than if you launch from a small hill ?

Tip: make a drawing!

My solution was to find the time taken for the projectile to reach the ground. Then sub that time, t, into the horizontal displacement formula vi cos Θ .t. Is there a reason why this is not viable?
 
  • #4
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The idea is excellent. But don't you think that h should appear somewhere in that time ?
You replace vyf by vi sin theta. That is not good. In the first place because vi sin theta is 0. In the second place
because vyf = gt.

Made a drawing yet ?
 
  • #5
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The idea is excellent. But don't you think that h should appear somewhere in that time ?
You replace vyf by vi sin theta. That is not good. In the first place because vi sin theta is 0. In the second place
because vyf = gt.

Made a drawing yet ?

I have the drawing already. It's really just the second half of the parabola.
 
  • #6
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The idea is excellent. But don't you think that h should appear somewhere in that time ?
You replace vyf by vi sin theta. That is not good. In the first place because vi sin theta is 0. In the second place
because vyf = gt.

Made a drawing yet ?

It's not an issue, is it? Since Θ = 0°, then, vi sin(0°) = 0.
 
  • #7
BvU
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Two approaches possible, in fact they are the same.
You can get vyf from an energy balance mgh = 1/2 m vyf^2
or you can get vyf by solving for y(t) = 0 the equation y(t) = y0 + vyi * t - 1/2 * g * t^2 an using that t in your expression for x.
 
  • #8
818
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Two approaches possible, in fact they are the same.
You can get vyf from an energy balance mgh = 1/2 m vyf^2
or you can get vyf by solving for y(t) = 0 the equation y(t) = y0 + vyi * t - 1/2 * g * t^2 an using that t in your expression for x.

It struck me that the equation y(t)= y(0) + vi.t -0.5gt^2 can be reduced to y(t) - y(0) = h = vi.t - 0.5gt^2. From there I can solve for time. I think this approach is a parallel reasoning to the one in my attempted solution and hence I do not think my answer in the OP is wrong. The only difference would be that using h allow me to express the solution in the intended manner the question demands.
 
  • #9
818
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Two approaches possible, in fact they are the same.
You can get vyf from an energy balance mgh = 1/2 m vyf^2
or you can get vyf by solving for y(t) = 0 the equation y(t) = y0 + vyi * t - 1/2 * g * t^2 an using that t in your expression for x.


h = vyi.t - 0.5gt^2
but since yf < yi , then, yf - yi = -h, and vyi = 0 since vi sin 0° = 0
hence, - h = -0.5gt^2
t = SQRT[2h/g] (time taken for projectile to reach the ground)

given time, t = SQRT [ 2h/g], the horizontal displace in this period of time
= vxi.t = vi cos(0°) [SQRT[2h/g]] = vi[SQRT[2h/g]]
 
  • #10
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Excellent.
 

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