Projectile motion

  • Thread starter Googl
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  • #1
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Hi all,

I have a question, where I am given just the angle at which the object is fired at to the horizontal and the amount of time the object stays in the air. How can I possibly find the distance at which the object will land from the origin if I don't have the velocity at which the object was fired? Or the highest point at which it will reach.

That is just impossible. Am I wrong?

Thanks.
 

Answers and Replies

  • #2
AlephZero
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Ignoring air resistance, the horizontal and vertical motion of the object are indepedent.

You know the time in the air, so you can find the vertical component of the velocity.

You also know the vertical component of the velocity is [itex]V\sin\theta[/itex] and the horizontal component is [itex]V\cos\theta[/itex]. You know [itex]\theta[/itex] so you can find [itex]V[/itex].
 
  • #3
2,745
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Divide the time by 2 to give you the time half way - the time to the peak.

You know acceleration = g, time = ttotal/2, and if you take the first part of the journey you know the final speed is zero (peak of journey, vertical velocity is zero).

So for the first part you've got vertical acceleration (g), final velocity (0) and time (ttotal/2).

From that you can plug those numbers into the equations of motion and get the maximum height reached and the initial velocity.
 
  • #4
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Ignoring air resistance, the horizontal and vertical motion of the object are indepedent.

You know the time in the air, so you can find the vertical component of the velocity.

You also know the vertical component of the velocity is [itex]V\sin\theta[/itex] and the horizontal component is [itex]V\cos\theta[/itex]. You know [itex]\theta[/itex] so you can find [itex]V[/itex].

Let's say angle is 50 degrees and time 20s

So you're saying;

[itex]V\cos\theta[/itex] = [itex]u\cos50[/itex] + axt
Horizontal acceleration is 0 (x-axis)
[itex]V\cos\theta[/itex] = 0 + 0x20s
[itex]V\cos\theta[/itex] = 0
V = cos50/0
V = 0
 
  • #5
AlephZero
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No, do what I said.

The initial vertical velocity = [itex]V \sin \theta[/itex]

The vertical acceleration = [itex]-g[/itex]
Vertical displacement = [itex]y = (V\sin\theta) t - g t^2 / 2[/itex]

When [itex]y = 0[/itex]

[itex]t = 0[/itex] or [itex]t = 2V\sin\theta /g[/itex]

[itex]V = gt / 2 \sin\theta[/itex]
 
  • #6
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Sorry, I am trying to find how far the object lands (that is in the x-axis).

Angle = 20 degress.
Time = 10s

The initial horizontal velocity = V cos20

The horizontal acceleration = 0
Horizontal displacement = y = V cos20 x 10 - 0 x 10^2 / 2

Is that right?
 
  • #7
Redbelly98
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Is that right?
Yes it is right, but it is useless if you don't know V. Do what AlephZero said to figure out what V is.
 
  • #8
2,745
22
Either I'm missing something here or you could plug the three values I pointed out into the SUVAT equations and had the answers out in seconds.

Given the trouble being had with AlephZeros method I'd at least have thought you'd give it a go.
 

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