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August 15th, 2015, 11:38 AM   #1
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count arrow flight distance

How to calculate how far does arrow fly when only thing known is starting speed of the arrow and air friction slowing down the arrow?

speed = 50 units
air friction = 10 units

I hope I posted this on right section, sorry my English isn't best when it comes to math.

Thanks.

Last edited by skipjack; August 15th, 2015 at 01:42 PM.
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August 15th, 2015, 04:31 PM   #2
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The angle of elevation also matters.
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August 16th, 2015, 12:54 AM   #3
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Quote:
Originally Posted by Azzajazz View Post
The angle of elevation also matters.
Not really I'm not doing simulation (yet) I just want a simple range for the arrow.

another example could be that car is driving 50 mp/h and braking so that it loses speed 10 mp/h every second. How long does the car go until it stops?
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August 16th, 2015, 04:49 AM   #4
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Obviously, the car stops after 5 seconds.
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August 16th, 2015, 05:01 AM   #5
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Quote:
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Obviously, the car stops after 5 seconds.
I did not ask how fast the car stops - but how far it will go until it stops.
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August 16th, 2015, 07:13 AM   #6
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You asked how long the car goes (until it stops), which is 5 seconds.

Its average speed (while braking) is 25mph, and 5 seconds is (1/720) hours, so the distance the car covers in that 5 seconds of braking is (25/720) miles, i.e. (5/144) miles (or 183 feet 4 inches if you prefer).
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August 16th, 2015, 07:52 AM   #7
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Try posting in the Differential Equations
section. [First search the phrase
'projectile with air resistance'.]
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August 17th, 2015, 05:19 PM   #8
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If the car (or arrow) is losing speed at a constant 10 mph (i.e. it's acceleration is -10 mph) then dv/dt= -10. Integrating, v= -10t+ C for some constant C. You said that the speed was initially v(0)= 50= -10(0)+ C so C= 50. That is, v= -10t+ 50 But v= dx/dt, where x is the distance moved so dx/dt= -10t+ 50. Now, integrating again, x= -5t^2+ 50t+ C with C, again, a "constant of integration". Since the question is about "distance moved" we can take x(0) to be 0 so that C= 0. x(t)= -5t^2+ 50t. The speed, v= -10t+ 50 will be 0 when -10t+ 50= 0, -10t= -50, t= 5 as skipjack said. Now, what is x(5)?
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August 18th, 2015, 02:52 AM   #9
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Math Focus: Physics, mathematical modelling, numerical and computational solutions
If acceleration is zero (no drag), use:

$\displaystyle x = vt + x_0$

If acceleration is constant, (drag is constant with speed) then use the SUVAT equations. For example, you could use

$\displaystyle s = ut + \frac{1}{2}at^2$

If acceleration is not constant (drag that varies with speed), then you need to solve a differential equation based on Newton's laws. A drag force proportional to $\displaystyle v$ or $\displaystyle v^2$ is typical:

$\displaystyle F = F_{drag} = \alpha v^2 = ma$

$\displaystyle a = \frac{\alpha}{m} v^2$

$\displaystyle \frac{d^2 x}{dt^2} - \frac{\alpha}{m} \left(\frac{dx}{dt}\right)^2 = 0$

A drag that varies with speed is the most realistic in terms of what happens to real objects, but this is also the most complicated to solve. Have fun!
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August 18th, 2015, 04:01 AM   #10
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Quote:
Originally Posted by Benit13 View Post

If acceleration is constant, (drag is constant with speed) then use the SUVAT equations. For example, you could use

$\displaystyle s = ut + \frac{1}{2}at^2$
this seems like what I need. but how do I know what at, ut and s stands for?
(s I believe is speed)
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