Shoot it!

Classical Mechanics Level 3

Let say you have an automated gun which can shoot the target by determining the horizontal and vertical distance to the target from the gun. It can also shoot at any velocity and at any angle. But you programmed the gun so that it will fire the bullet at minimum velocity possible to hit the target.

During a test run your gun found that the target is situated at $70m$ height and at $140m$ horizontal distance. Find the velocity (in $m/s$ ) your gun is going to fire at.

Note:

Imagine the bullet trajectory to be projectile motion.
Take value of $g$ as $10m/s^2$ .
Also as usual ignore air resistance and other real life factors.

The answer is 47.5946.

2 solutions

A Former Brilliant Member
Aug 4, 2020

The equation of trajectory of the projectile is

$y=x\tan α-\dfrac {gx^2\sec^2 α}{2u^2}$

$\implies \tan^2 α-\dfrac {2u^2}{gx}\tan α+1+\dfrac {2u^2y}{gx^2}=0$

Since $α$ is real, therefore $u^4\geq g^2x^2+2gyu^2$

$\implies u\geq \sqrt {g\left (y+\sqrt {x^2+y^2}\right )}$

$=\sqrt {700(\sqrt 5+1)}\approx \boxed {47.5946}$ m/s.

Steven Chase
Aug 1, 2020

I originally solved and got the expected answer by assuming that the vertical velocity is zero when the projectile strikes the target. However, this is not the optimal solution. A parameter sweep shows that the minimum launch speed is $47.59$ , and it occurs when the launch velocity is $(v_x, v_y) = (25.02, 40.49)$ .

import math

H = 70.0    # height
D = 140.0   # horizontal distance
g = 10.0    # gravity

dvx = 10.0**(-3.0)  # horizontal velocity increment

vx = dvx

vmin = 99999999.0

while vx <= 100.0:  # sweep horizontal velocity over a range

    tf = D/vx  # final time

    # H = vy*tf - 0.5*g*(tf**2.0)   # vertical kinematic equation

    vy = (H + 0.5*g*(tf**2.0))/tf   # solve for vertical velocity

    v = math.hypot(vx,vy)        # launch speed

    if v < vmin:     # keep track of min launch speed
        vmin = v
        vx_store = vx
        vy_store = vy

    vx = vx + dvx

print vmin
print vx_store
print vy_store

#>>> 
#47.5946171773
#25.022
#40.4863816641
#>>>

@Steven Chase in first attempt I got 47.5 , but it was incorrect.
After some time i was thinking that you will delete you solution, but you just changed it according to correct answer. Smart guy.

Talulah Riley - 10 months, 2 weeks ago

Yeah, I was a bit careless with the first solution

Steven Chase - 10 months, 2 weeks ago

@Steven Chase but your strategy of thinking that the Particle have vertical velocity zero at that point is incorrect. Even I don't know what is incorrect in that method.
I will think of it. Interesting!

Talulah Riley - 10 months, 2 weeks ago

@Talulah Riley – I have now posted an entirely different solution which gets the right answer

Steven Chase - 10 months, 2 weeks ago

@Steven Chase – @Steven Chase But what is wrong in your previous method.??

Talulah Riley - 10 months, 2 weeks ago

@Talulah Riley – It is based on intuition, and intuition turns out to be wrong in this case. The actual optimization problem is more complex

Steven Chase - 10 months, 2 weeks ago

@Steven Chase – @Steven Chase so the conclusion is if we want to through a particle to a particular coordinate it doesn't mean that, at that coordinate, the particle will have $v_{y}=0,v_{x}=v \cos \theta$ .
Extreme cases may exist.

Talulah Riley - 10 months, 2 weeks ago

@Talulah Riley – That's correct, the horizontal and vertical velocities are both nonzero at that point

Steven Chase - 10 months, 2 weeks ago

@Sanjay Swain Yes, the answer is approximately 47.595 m/s. Your problem answer is incorrect.

The trajectory is:

$\displaystyle y = \tan\theta x - \frac{1}{2}g\frac{x^2}{v^2 \cos^2 \theta}$

It follows that the velocity $v$ is a function of $\theta$ parameterized by $x$ and $y$ :

$\displaystyle v( \theta ) = \left( \frac{ gx^2 }{ 2 \cos^2 \theta ( y - \tan\theta x ) } \right)^\frac{1}{2}$

An optimization of $v (\theta)$ is found by taking the derivative of $v$ w.r.t. $\theta$ and equating it to $0$

$\displaystyle \frac{ dv }{ d \theta } = \frac{d}{d \theta} \left( \frac{ gx^2 }{ 2 \cos^2 \theta ( y - \tan\theta x ) } \right)^{\frac{1}{2}} = 0$

Once you find $\theta_{opt.}$ ( confirm you have a minimum by second derivative tests or a simple plot of $v( \theta )$ , substitute back into $v ( \theta )$ ).

$\displaystyle v ( \text { 1.017} ) \approx \text{ 47.595 } \frac{ \text{m} }{ \text{s} }$

Eric Roberts - 10 months, 2 weeks ago

Shoot it!

The answer is 47.5946.

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