Distance problem, Using Launcher need to find objects distance

  • Thread starter crazydude032
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In summary, the conversation is about finding the distance of an object without actually shooting from a launcher. The initial velocity is 518.28 feet per second, the final velocity is 0 ft/s, gravity's constant is -32.15 feet/sec2, and the object is launched at a 45 degree angle from 2.12 feet off the ground. The formula to find the distance is Δx=viΔt, and after solving for time, the distance is approximately 8392.392 feet.
  • #1
crazydude032
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Hows it going guys, I need to now how to find distance of an object without actually shooting from my launcher. I have that it shoots 518.28 feet per sec ( initial velocity) ends at 0 ft/s (final velocity), that gravity's constant is -32.15 feet/sec2(9.8 meters/sec2), I also have that it shoots at a 45 degree angle and it launches 2.12 feet off the ground, objects weight is .0154 lbs. It feels like I have everything I need to find distance but I keep getting huge numbers. If you could solve this for me or just point me to the formulas that I would need I would greatly appreciate it:-p , This is going to help launch a projectile for a project and this formula will save me a lot of time and money
Thanx again
 
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  • #2
crazydude032 said:
...It feels like I have everything I need to find distance but I keep getting huge numbers.
...If you could solve this for me or just point me to the formulas

So, you obviously used some formula to get these huge numbers, so, present your work. :cool:
 
  • #3
Here are the formulas
ΔX= 1/2 (initial velocity in x direction + final velocity) Δtime
vfx= vix+accelerationΔt
Δx=viΔt+ 1/2 a (Δt)^2
vf^2=vi^2+2aΔX

vf= final velocity
vi=initial velocity
a= acceleration
t=time
Δ= displacement
You must remember that in an equation vi and vf must be either along the X axis or along the Y axis.

I solved it too
Each equation can be applied to either the x or the y axis.
that means
Δx= vit+1/2at^2 is in the x direction
Δy=vit+1/2at^2 is in the y direction

Here it is solved
vi in x direction = 518.28 cos 45
vi in y direction = 518.28 sin 45

Δy=viΔt+1/2aΔt^2
-2.12= 366.48Δt-16t^2
16t^2-366.48t-2.12=0
Solve for t
t= 22.9 seconds
t in y direction = t in x direction

since acceleration = 0 in x direction
Δx=viΔt
Δx=366.48*22.9
Δx ~ 8392.392 ft

That is a long distance. If it's way of, I'm sure there is a problem with you initial velocity. 518.28 ft/s is too much.
 

Related to Distance problem, Using Launcher need to find objects distance

1. How does the launcher calculate the distance of an object?

The launcher uses a combination of the object's initial velocity and the time it takes for the object to reach its peak height to calculate the distance.

2. Can the launcher be used to measure the distance of any object?

Yes, as long as the object can be launched and its initial velocity and peak height can be measured accurately, the launcher can be used to calculate its distance.

3. What factors can affect the accuracy of the distance measurement using the launcher?

The accuracy of the distance measurement can be affected by factors such as air resistance, wind, and the precision of the measurements of initial velocity and peak height.

4. Is there a minimum or maximum distance that the launcher can measure?

No, the launcher does not have a maximum or minimum distance limit. However, the accuracy of the measurement may decrease as the distance increases.

5. Can the launcher be used to measure the distance of objects in different environments?

Yes, the launcher can be used to measure the distance of objects in various environments as long as the factors mentioned in question 3 are taken into consideration.

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