Don't Ignore Air Resistance!

Classical Mechanics Level 5

A projectile with mass m m is thrown from position ( 0 , 0 ) \left(0,0\right) with initial velocity u u upwards at an angle θ \theta with the x-axis.

After being thrown, there are two forces acting on the particle. There is a constant gravitational acceleration acting downwards with magnitude g g . There is also acceleration due to air resistance, acting against the direction of the projectile's instantaneous velocity and having magnitude equal to k v m \frac{kv}{m} , where v v is the instantaneous velocity of the projectile. Note that g g and k k are assumed to be constant.

Find the projectile's vertical displacement y y in terms of the projectile's horizontal displacement x x .

If your answer is of the form

y ( x ) = x A ( tan ⁡ ( θ ) + B g m k u cos ⁡ ( θ ) + C ) + g D m E k F ln ⁡ ( 1 − G k x m u cos ⁡ ( θ ) ) , y(x)=x^A\left(\tan\left(\theta\right)+\frac{Bgm}{ku\cos\left(\theta\right)}+C\right)+\frac{g^Dm^E}{k^F}\ln\left(1-\frac{Gkx}{mu\cos\left(\theta\right)}\right),

where A , B , C , D , E , F , G A,B,C,D,E,F,G are non-negative integers, enter A + B + C + D + E + F + G A+B+C+D+E+F+G .

Bonus:

According to the above simple model, given a fixed initial velocity u u , does throwing at an angle of 4 5 ∘ 45^\circ always give you maximum horizontal displacement when the projectile crosses the x-axis?

If you enjoyed this problem, you might also want to try this one .


The answer is 8.

Nick Turtle by Nick Turtle , 21, USA

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1 solution

Mark Hennings
May 12, 2018

We have the vector differential equation m r ¨ = − k r ˙ − m g j m \ddot{\mathbf{r}} \; = \; -k\dot{\mathbf{r}} - mg\mathbf{j} so that x ¨ + k m x ˙ = 0 y ¨ + k m y ˙ = − g \ddot{x} + \tfrac{k}{m}\dot{x} \; = \; 0 \hspace{2cm} \ddot{y} + \tfrac{k}{m}\dot{y} \; = \; -g The first integrals of these equations are x ˙ e k t / m = u cos ⁡ θ y ˙ e k t / m = − m g k ( e k t / m − 1 ) + u sin ⁡ θ \dot{x}e^{kt/m} \; = \; u\cos\theta \hspace{2cm} \dot{y}e^{kt/m} \; = \; -\tfrac{mg}{k}(e^{kt/m}-1) + u\sin\theta Integrating one more time gives x = m u k cos ⁡ θ ( 1 − e − k t / m ) y = m k ( u sin ⁡ θ + m g k ) ( 1 − e − k t / m ) − m g k t x \; = \; \tfrac{mu}{k}\cos\theta(1 - e^{-kt/m}) \hspace{2cm} y \; = \; \tfrac{m}{k}(u\sin\theta + \tfrac{mg}{k})(1 - e^{-kt/m}) - \tfrac{mg}{k}t and hence we obtain the equation y = ( tan ⁡ θ + m g k u cos ⁡ θ ) x + m 2 g k 2 ln ⁡ ( 1 − k x m u cos ⁡ θ ) y \; = \; \big(\tan\theta + \tfrac{mg}{ku\cos\theta}\big)x + \tfrac{m^2g}{k^2}\ln\big(1 - \tfrac{kx}{mu\cos\theta}\big)

This makes the answer should 1 + 1 + 0 + 1 + 2 + 2 + 1 = 8 1 + 1 + 0 + 1 + 2 + 2 + 1 = \boxed{8} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

"No" The optimized angle for achieving maximum horizontal range is not 45 degrees As asked in the bonus

Suhas Sheikh - 3 years ago

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