Distance problem, Using Launcher need to find objects distance

[crazydude032]

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:tongue: , 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

 

[radou]

...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.

 

[apples]

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.