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April 30th, 2014, 08:49 PM  #1 
Member Joined: Apr 2014 From: Perth Posts: 34 Thanks: 0  Differential equation that involve projectile motion ;w;
Any expert know how to do this ? differential equation : When an object is thrown up vertically, the time taken to reach the highest point and the time taken to fall back to the initial projection can be compared by constructing a mathematical model using a differential equation to describe the phenomenom. An object of mass m kg is projected vertically upward from the ground with initial velocity u ms1. Assume that the forces acting on the object are the gravitational force and the retarding forces due to air resisitance with a magnitude of α│v│, where α is a positive constant and v is the velocity of the object at time t. When an object is thrown up vertically, the time taken to reach the highest point and the time taken to fall back to the initial projection can be compared by constructing a mathematical model using a differential equation to describe the phenomenom. An object of mass m kg is projected vertically upward from the ground with initial velocity u ms1. Assume that the forces acting on the object are the gravitational force and the retarding forces due to air resisitance with a magnitude of α│v│, where α is a positive constant and v is the velocity of the object at time t. m dv/dt=∝vmg (a)Express v in term of t. (b)Find the height s in term of t. (c) If the mass of the object is 0.1 kg, the initial velocity is 5 ms1 , α is 1.0 kgms1 and g is 10ms2. i)Calculate the time taken for the object to reach the highest point ii)Estimate the time taken for the object to descend from the highest point to the ground. 
May 1st, 2014, 03:16 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Part (a): $\displaystyle m\frac{dv}{dt}=\alpha v  mg$ $\displaystyle \frac{dv}{dt}=\frac{\alpha v}{m}  g$ $\displaystyle \frac{dv}{dt}+\frac{\alpha v}{m}= g$ This is a firstorder ODE that can be solved using the integrating factor method. Let the integrating factor $\displaystyle P(t) = \exp\left(\int\frac{\alpha}{m}dt\right)$ $\displaystyle P(t) = \exp\left(\frac{\alpha t}{m}\right)$ $\displaystyle \frac{d}{dt}\left(vP(t)\right) = g\left(P(t)\right)$ $\displaystyle \frac{d}{dt}\left(v\exp\left(\frac{\alpha t}{m}\right)\right) = g\left(\exp\left(\frac{\alpha t}{m}\right)\right)$ $\displaystyle v\exp\left(\frac{\alpha t}{m}\right) = \int g\exp\left(\frac{\alpha t}{m}\right)dt$ $\displaystyle v\exp\left(\frac{\alpha t}{m}\right) = \frac{mg}{\alpha}\exp\left(\frac{\alpha t}{m}\right) + C$ $\displaystyle v = C\exp\left(\frac{\alpha t}{m}\right)  \frac{mg}{\alpha}$ We can eliminate the $\displaystyle C$ using the boundary condition that when $\displaystyle t=0, v=u$. $\displaystyle C = u + \frac{mg}{\alpha}$ $\displaystyle v = \left(u+\frac{mg}{\alpha}\right)\exp\left(\frac{\alpha t}{m}\right)  \frac{mg}{\alpha}$ Part (b) $\displaystyle s = \int vdt$ $\displaystyle s = \int\left[\left(u+\frac{mg}{\alpha}\right)\exp\left(\frac{\alpha t}{m}\right)  \frac{mg}{\alpha}\right]dt$ $\displaystyle s = \frac{m}{\alpha}\left(u+\frac{mg}{\alpha}\right) \exp\left(\frac{\alpha t}{m}\right)  \frac{mgt}{\alpha} + C$ We can eliminate the $\displaystyle C$ using the boundary condition that when $\displaystyle t=0, s=0$. $\displaystyle C = \frac{m}{\alpha}\left(u+\frac{mg}{\alpha}\right)$ $\displaystyle s = \frac{m}{\alpha} \left(u+\frac{mg}{\alpha}\right) \left[1\exp\left(\frac{\alpha t}{m}\right)\right]  \frac{mgt}{\alpha}$ Part (c) $\displaystyle m = 0.1 kg$ $\displaystyle u = 5 m/s$ $\displaystyle \alpha = 1 kgm/s$ $\displaystyle g = 10m/s^2$ Part i) The object reaches its maximum height when its first derivative is zero. since the first derivative is its velocity, which we have already calculated, we just set $\displaystyle v=0$ and rearrange for $\displaystyle t$. $\displaystyle \left(u+\frac{mg}{\alpha}\right)\exp\left(\frac{\alpha t}{m}\right)  \frac{mg}{\alpha} = 0$ $\displaystyle \left(u+\frac{mg}{\alpha}\right)\exp\left(\frac{\alpha t}{m}\right) = \frac{mg}{\alpha}$ $\displaystyle \exp\left(\frac{\alpha t}{m}\right) = \frac{\frac{mg}{\alpha}}{u + \frac{mg}{\alpha}}$ $\displaystyle \exp\left(\frac{\alpha t}{m}\right) = \frac{mg}{u\alpha + mg}$ $\displaystyle t = \frac{m}{\alpha}\ln\left(\frac{mg}{u\alpha + mg}\right)$ $\displaystyle t = \frac{m}{\alpha}\ln\left(\frac{u\alpha + mg}{mg}\right)$ Inserting numbers gives $\displaystyle t = 0.1\ln6 = 0.179s$ Part ii) The object hits the ground when $\displaystyle s = 0$ $\displaystyle \frac{m}{\alpha} \left(u+\frac{mg}{\alpha}\right) \left[1\exp\left(\frac{\alpha t}{m}\right)\right]  \frac{mgt}{\alpha} = 0$ $\displaystyle gt = \left(u+\frac{mg}{\alpha}\right) \left[1\exp\left(\frac{\alpha t}{m}\right)\right]$ This cannot be solved as it isn't possible to separate $\displaystyle t$ out of the equation. However, an estimate can be found using iteration: $\displaystyle t_{n+1} = \left(\frac{u}{g}+\frac{m}{\alpha}\right) \left[1\exp\left(\frac{\alpha t_n}{m}\right)\right]$ Plugging numbers in gives a much nicer formula to work with $\displaystyle t_{n+1} = 0.6 \left( 1  \exp \left(10t_n\right) \right)$ With an initial guess of $\displaystyle t_0 = 0.4$, the iteration above converges rapidly to $\displaystyle t = 0.59849012264$, accurate to 11 decimal places. Let me know if there any typos or errors! 
May 1st, 2014, 03:25 AM  #3 
Senior Member Joined: Dec 2012 Posts: 372 Thanks: 2 
$\displaystyle m \frac{dv}{dt} = \alpha v  mg \\ $ Integrating both sides with respect to t, you get $\displaystyle m v =  \alpha s  mgt + k$ where $\displaystyle s$ is the distance covered, and $\displaystyle k$ is a constant of integration. However, we may also integrate the equation as follows $\displaystyle m\left( \frac{dv}{\alpha v + mg} \right) = dt \\ \Rightarrow \frac{m}{\alpha}ln\alpha v + mg = t + c \\ \Rightarrow v = \frac{[\pm \exp (\frac{\alpha}{m} (t + c))  mg]}{\alpha} $ From the first integration, you get $\displaystyle s = \frac{1}{\alpha}(mv  mgt + k) = \frac{1}{\alpha}\left( \frac{m[\mp \exp (\frac{\alpha}{m} (t + c)) + mg]}{\alpha} mgt + k \right)$ Using your initial conditions, I compute my constant of integration $\displaystyle c$ to equal 0.1792, and since at the highest point $\displaystyle v = 0$ then the time taken to reach this point is easily derived to be $\displaystyle 0.1792 s.$ I estimate that this should be the same time taken to get from the peak back to the launching point. 
May 1st, 2014, 09:01 PM  #4 
Member Joined: Apr 2014 From: Perth Posts: 34 Thanks: 0 
Sir Benit13 , thank you so much , I understand for the most part except for the last one . Can you further explain the iteration method ? As for the initial guess , is that mean I can assume/input any number for it other than 0.4 ? >.< And is it possible for me to use NewtonRaphson method instead of iteration ?? AfroMike , thanks for your reply 
May 1st, 2014, 10:20 PM  #5 
Member Joined: Apr 2014 From: Perth Posts: 34 Thanks: 0 
Ohhhh and for the first part , how come the exp(−αtm) suddenly have negative sign in it ?? my maths is really weak , I'm sorry to had trouble you T.T 
May 2nd, 2014, 02:25 AM  #6  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
No worries! Quote:
You can use any initial guess and, provided $\displaystyle f(x)$ converges, you will reach a solution. It should be noted however that whether or not an iteration scheme will converge to a particular solution depends on your initial guess (in some circumstances), so it is good practise to pick something sensible near where you think the answer could be. Any method that can be used to find a numerical solution to find the roots of an equation, such as NewtonRaphson, can be used Quote:
 
May 3rd, 2014, 08:16 AM  #7 
Member Joined: Apr 2014 From: Perth Posts: 34 Thanks: 0 
well explained , thank you sir *bow* And I kinda try question c with a table when the value of alpha is different while others remain the same to get the time taken for i and ii , surprisingly my calculation for the time taken is decreasing where it suppose to be increasing from what I had read about air resistance sobs why is it like that ?? OTL unless i calculate it wrongly , and I used NewtonRaphson method with it 
May 6th, 2014, 08:48 AM  #8 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
If you increase the coefficient of the air resistance, you might expect the time taken to hit the floor to be longer because the drag is larger and the object will not accelerate as much, but you also have to consider the fact that the height the object reaches is smaller, so the duration spent in the air will decrease because of the reduced distance to travel. These two effects will compete, so a careful analysis will have to be made to determine the final trend (i.e. an increase or decrease in airborne duration with increasing air resistance is possible. It depends on the specific system). In addition, real physics of air resistance is actually conducted with a force term that is proportional to $\displaystyle v^2$ rather than $\displaystyle v$. This makes air resistance much more important for the motion of an object than your studies may perhaps lead you to believe, although a good physics teacher will always mention where it is or isn't appropriate to neglect it. I guess this is why so much money is spent on making vehicles streamlined so that energy isn't wasted accelerating an object that suffers from significant drag. Also, an exam question involving a term with $\displaystyle v^2$ would be incredibly evil as it is nonlinear and therefore a bugger to solve. In any case, using a term directly proportional to $\displaystyle v$ is probably a fair approximation. 

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