Finding Distance w/o Velocity or Max Height

In summary, AlephZero was giving a method for finding the distance an object will land from its origin, but this method is useless without knowing the initial velocity.
  • #1
Googl
111
1
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.
 
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  • #2
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
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
AlephZero said:
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
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
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
Googl said:
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
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.
 

1. How can distance be calculated without knowing velocity or maximum height?

Distance can be calculated using the formula d = (1/2)gt^2, where g is the acceleration due to gravity and t is the time taken to reach the ground. This formula assumes that the object starts and ends at the same height.

2. Is this formula applicable for all objects?

No, this formula is only applicable for objects in a uniform gravitational field, such as on Earth. For objects in non-uniform gravitational fields or in space, other equations may need to be used.

3. Can this formula be used to find distance for objects thrown at an angle?

Yes, for objects thrown at an angle, the initial velocity can be broken down into horizontal and vertical components. The formula can then be applied to the vertical component to find the distance traveled in the vertical direction.

4. Is there a more accurate way to calculate distance without knowing velocity or maximum height?

Yes, if the object's motion is known to be affected by air resistance, a more complex formula taking into account air resistance can be used. However, this requires additional information such as the object's mass, cross-sectional area, and air density.

5. How does changing the acceleration due to gravity affect the calculated distance?

Changing the acceleration due to gravity will directly affect the calculated distance. For example, if the object is on a different planet with a different gravity, the value for g in the formula will need to be substituted with the appropriate value for that planet.

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