My Math Forum count arrow flight distance

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 August 15th, 2015, 11:38 AM #1 Newbie   Joined: Aug 2015 From: finland Posts: 9 Thanks: 0 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.
 August 15th, 2015, 04:31 PM #2 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 The angle of elevation also matters.
August 16th, 2015, 12:54 AM   #3
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Quote:
 Originally Posted by Azzajazz 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?

 August 16th, 2015, 04:49 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,128 Thanks: 2336 Obviously, the car stops after 5 seconds.
August 16th, 2015, 05:01 AM   #5
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Quote:
 Originally Posted by skipjack 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.

 August 16th, 2015, 07:13 AM #6 Global Moderator   Joined: Dec 2006 Posts: 21,128 Thanks: 2336 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).
 August 16th, 2015, 07:52 AM #7 Senior Member   Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 Try posting in the Differential Equations section. [First search the phrase 'projectile with air resistance'.]
 August 17th, 2015, 05:19 PM #8 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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)?
 August 18th, 2015, 02:52 AM #9 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 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!
August 18th, 2015, 04:01 AM   #10
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Quote:
 Originally Posted by Benit13 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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