Maximize Range of Projectile Launched from Cliff

[DeltaForce]

Homework Statement

Find the expression for an angle x that maximizes range of the projectile launched from a cliff of height h.

Relevant Equations

t1 = 2vsin(x)/g
h = vsin(x)(t_2) + 0.5g(t_2)^2
t2 = (-vsin(x) + sqrt((vsinx)^2+2gh))/g
R = v_x*(t_1 + t_2)

I found the function of the range of the projectile launched from a cliff.

R = vcos(x)((vsin(x))+sqrt(v^2sin^2(x)+2gh))/g

I stopped here because I feel like taking the derivative and optimizing for maximum would spiral out of control.

Is there another approach to this problem?

 

[berkeman]

Homework Statement:: Find the expression for an angle x that maximizes range of the projectile launched from a cliff of height h.
Relevant Equations:: t1 = 2vsin(x)/g
h = vsin(x)(t_2) + 0.5g(t_2)^2
t2 = (-vsin(x) + sqrt((vsinx)^2+2gh))/g
R = v_x*(t_1 + t_2)

I stopped here because I feel like taking the derivative and optimizing for maximum would spiral out of control.

Is there another approach to this problem?

I'm not able to do much with your text equations (see the LaTeX Guide in the lower left of the edit window), but I think the right approach is to differentiate the range versus the angle, etc. Why would you think a different approach is available and better?

 

[haruspex]

Homework Statement:: Find the expression for an angle x that maximizes range of the projectile launched from a cliff of height h.
Relevant Equations:: t1 = 2vsin(x)/g
h = vsin(x)(t_2) + 0.5g(t_2)^2
t2 = (-vsin(x) + sqrt((vsinx)^2+2gh))/g
R = v_x*(t_1 + t_2)

I found the function of the range of the projectile launched from a cliff.

R = vcos(x)((vsin(x))+sqrt(v^2sin^2(x)+2gh))/g

I stopped here because I feel like taking the derivative and optimizing for maximum would spiral out of control.

Is there another approach to this problem?

It is unnecessary to split it into two time periods. The same equation applies throughout the trajectory.
It can be simpler to differentiate the individual equations instead of combining them into the form R=... first. You will get terms like t', being the derivative of t wrt the angle, but these can be eliminated later.

 

[DeltaForce]

I'm not able to do much with your text equations (see the LaTeX Guide in the lower left of the edit window), but I think the right approach is to differentiate the range versus the angle, etc. Why would you think a different approach is available and better?

Sorry about the messy equations. When I took the derivative of range vs angle, I got a very complicated output. Therefore I thought it was difficult to optimize using that solution to find an expression for the angle. That's why I thought there may be a different or better approach available.

 

[DeltaForce]

It is unnecessary to split it into two time periods. The same equation applies throughout the trajectory.
It can be simpler to differentiate the individual equations instead of combining them into the form R=... first. You will get terms like t', being the derivative of t wrt the angle, but these can be eliminated later.

Can you show me how you would do this problem? You seem like you know what's up.

 

[phinds]

Can you show me how you would do this problem? You seem like you know what's up.

You've been given a strong hint. How about you show some effort before asking for more help?

 

[DeltaForce]

You've been given a strong hint. How about you show some effort before asking for more help?

I came up with this problem and did the work, that is hardly an example of low-effort. I don't think I understand the hint well. Do you happen to know the solution?

 

[haruspex]

Can you show me how you would do this problem? You seem like you know what's up.

First, just use one time period, t, and write the displacement equations using that. Post what you get.

 

[DeltaForce]

First, just use one time period, t, and write the displacement equations using that. Post what you get.

Thanks for your replies. I did that. But I got the same displacement equation that I posted.
t = (vsin(x)+sqrt(v^2*sin^2(x)+2gh))/g. I got this for the time period.

 

[haruspex]

Thanks for your replies. I did that. But I got the same displacement equation that I posted.
t = (vsin(x)+sqrt(v^2*sin^2(x)+2gh))/g. I got this for the time period.

No, I didn't mean to obtain an expression for the time. Just write the two usual (SUVAT) displacement equations, R=..., h=... in terms of v, t, x, g.

 

[DeltaForce]

No, I didn't mean to obtain an expression for the time. Just write the two usual (SUVAT) displacement equations, R=..., h=... in terms of v, t, x, g.

Like R = vcos(x)t and
H = vsin(x)t - 1/2gt^2?

 

[haruspex]

Like R = vcos(x)t and
H = vsin(x)t - 1/2gt^2?

Yes.
Now differentiate them wrt x. Remember that t is a function of x, so the derivatives will involve dt/dx terms, which you can abbreviate to t'.

Edit: @kuruman's trick, below, is very neat, but you might find it useful to continue to explore the path I have proposed as a general technique for handling messy optimisations.

 

[kuruman]

Is there another approach to this problem?

There is and you don't have to solve a quadratic or take derivatives. The horizontal range can be written as $$R=\frac{|\vec v_0 \times \vec v_{\!f}|}{g}$$ where $\vec v_0$ and $\vec v_{\!f}$ are, respectively, the initial and final velocity vectors. This is maximum when the two vectors are perpendicular in which case the numerator becomes the product of the initial and final speeds. You can find the final speed given the initial speed and the vertical displacement in a number of ways, e.g kinematic equations, conservation of energy, etc.

For an explanation see
Source https://www.physicsforums.com/insights/how-to-master-projectile-motion-without-quadratics/

Sorry for diverting the thread to a new direction, but this is conceptually easier to understand and quicker to implement.

 

[DeltaForce]

This is an amazing solution. Thanks for taking the time to write this. The source you linked also looked fantastic. Thanks!

 

[PeroK]

This is an amazing solution. Thanks for taking the time to write this. The source you linked also looked fantastic. Thanks!

I agree with @haruspex. Solving the quadratic equation is fundamentally important and useful across all of physics at all levels. You need to master the techniques involved.

 

[DeltaForce]

Yes.
Now differentiate them wrt x. Remember that t is a function of x, so the derivatives will involve dt/dx terms, which you can abbreviate to t'.

Edit: @kuruman's trick, below, is very neat, but you might find it useful to continue to explore the path I have proposed as a general technique for handling messy optimisations.

I'm not that familiar with this strategy of differentiation. But I thank you for giving me this new technique to me. I definitely have to try this out, and hopefully master it.

 

[DeltaForce]

Yes.
Now differentiate them wrt x. Remember that t is a function of x, so the derivatives will involve dt/dx terms, which you can abbreviate to t'.

Edit: @kuruman's trick, below, is very neat, but you might find it useful to continue to explore the path I have proposed as a general technique for handling messy optimisations.

Okay I have derived both equations.

dR/dx = v*(cos(x)t' - sin(x)t)
dH/dx = v*(cos(x)t + sin(x)t') - gt*t'

I am unsure where to take these results. Can you help me?

 

[haruspex]

Okay I have derived both equations.

dR/dx = v*(cos(x)t' - sin(x)t)
dH/dx = v*(cos(x)t + sin(x)t') - gt*t'

I am unsure where to take these results. Can you help me?

Since we are maximising R, and H is a constant, those derivatives are zero.
The first of the above equations then leads to $t'/t=\tan(x)$.
Substituting in the second gives $v=gt'\cos(x)$.
So now you can express t and t' in terms of v, g, x.
Substituting for those in the equation for H should get you nearly there.