Optimal Trajectory Angle for Maximum Area?

[Phantoful]

Homework Statement

A ball is thrown at an initial speed v from level ground. What angle θ should be chosen so that the area under the trajectory is maximized?

Homework Equations

d = V_ot - (1/2)gt²
Vt = d
Integration, derivatives, and trigonometry

The Attempt at a Solution

I've tried to find an equation for the curve by saying that x(t) = vtcos(theta) and y(t) = vtsin(theta) - (1/2)gt², but I'm not sure if that will get me anywhere, since I'm at a mental deadend on this part. Is there a right way to approach this problem?

 

[kuruman]

... but I'm not sure if that will get me anywhere ...

It will get you somewhere. Just find an expression for the horizontal distance in terms of v₀ and θ and then maximize that expression.

 

[Phantoful]

It will get you somewhere. Just find an expression for the horizontal distance in terms of v₀ and θ and then maximize that expression.

Wouldn't that just be v₀cos(θ)*t = d_x?

 

[kuruman]

Wouldn't that just be v₀cos(θ)*t = d_x?

This expression gives you the horizontal position at any time t. You don't want just any time, you want the specific time at which the projectile lands on the ground. How can you find that?

 

[Phantoful]

This expression gives you the horizontal position at any time t. You don't want just any time, you want the specific time at which the projectile lands on the ground. How can you find that?

The only way I would think is to set y(t) to zero and solve for t, then put it into the the x equation. Is this the right way?

 

[kuruman]

The only way I would think is to set y(t) to zero and solve for t, then put it into the the x equation. Is this the right way?

Yes. You will solve a quadratic equation that has two solutions indicating that the projectile is at zero height twice, once when it's launched and once when it lands.

 

[Phantoful]

Yes. You will solve a quadratic equation that has two solutions indicating that the projectile is at zero height twice, once when it's launched and once when it lands.

Ok, I put it in the equation and now time has been eliminated. I got (2V_o²*sinθ*cosθ)/g = d_x How would I find the maximum area under the curve, still? It's not clicking for me how I can link this to the area. It's not a max range question if that's what you meant

 

[kuruman]

Can you find an equation for the parabola now that you have the two points where y = 0? You will also need the maximum height which you can find in a number of ways. Then you can do the integral to find the area. Hint: A parabola in the form y(x) = C x (x-x₀) is zero at x = 0 and x = x₀. You need to find the appropriate values for C and x₀ that match the current situation.

 

[Phantoful]

Can you find an equation for the parabola now that you have the two points where y = 0? You will also need the maximum height which you can find in a number of ways. Then you can do the integral to find the area. Hint: A parabola in the form y(x) = C x (x-x₀) is zero at x = 0 and x = x₀. You need to find the appropriate values for C and x₀ that match the current situation.

Never mind, I got it! I did an integral of y(t)*x'(t) from 0 to the t you told me to find, for the answer (with respect to dt);

 

[kuruman]

What angle did you get? (Just checking)

 

[Phantoful]

What angle did you get? (Just checking)

pi/3