Optimal Trajectory Angle for Maximum Area?

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SUMMARY

The optimal trajectory angle for maximizing the area under the curve of a projectile launched from level ground is determined to be π/3 radians. The discussion highlights the use of kinematic equations, specifically the horizontal distance equation \(dx = v_0 \cos(\theta) t\) and the vertical position equation \(y(t) = v_0 \sin(\theta) - \frac{1}{2}gt^2\). Participants emphasized the importance of solving for the time when the projectile lands and integrating to find the area under the trajectory. The final solution involves deriving the area using the integral of the trajectory equations.

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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 = Vot - (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?
 
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Phantoful said:
... 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.
 
kuruman said:
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 = dx?
 
Phantoful said:
Wouldn't that just be v0cos(θ)*t = dx?
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?
 
kuruman said:
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?
 
Phantoful said:
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.
 
kuruman said:
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 (2Vo2*sinθ*cosθ)/g = dx 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
 
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.
 
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kuruman said:
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);
 
  • #10
What angle did you get? (Just checking)
 
  • #11
kuruman said:
What angle did you get? (Just checking)
pi/3
 

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