Optimal Trajectory Angle for Maximum Area?

In summary, To find the angle that maximizes the area under the trajectory of a ball thrown from level ground, one must find an expression for the horizontal distance and set it equal to zero to solve for the time at which the projectile lands. This time can then be used to find the maximum height and the appropriate values for C and x0 in the equation y(x) = C x (x-x0). Finally, the integral of y(t)*x'(t) from 0 to the time found will give the desired angle for maximum area.
  • #1
Phantoful
30
3

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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  • #2
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.
 
  • #3
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?
 
  • #4
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?
 
  • #5
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?
 
  • #6
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.
 
  • #7
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
 
  • #8
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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  • #9
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
 

FAQ: Optimal Trajectory Angle for Maximum Area?

1. What is the concept of maximizing area of trajectory?

The concept of maximizing area of trajectory involves finding the optimal path or trajectory for an object in motion that will cover the largest possible area. This is often used in fields such as physics, engineering, and mathematics to solve problems related to projectile motion, orbital mechanics, and optimization.

2. How is the area of trajectory calculated?

The area of trajectory can be calculated using the formula A = ½ bh, where A is the area, b is the base of the trajectory, and h is the height of the trajectory. The base and height can be determined using mathematical equations and principles such as calculus and trigonometry.

3. What factors affect the area of trajectory?

The area of trajectory is affected by various factors such as the initial velocity, angle of projection, air resistance, and gravitational pull. These factors can be manipulated to maximize the area of trajectory by adjusting the launch conditions or trajectory parameters.

4. Why is maximizing area of trajectory important?

Maximizing area of trajectory is important in many applications, such as in sports where athletes need to throw or hit an object as far as possible, or in space exploration where spacecrafts need to cover the largest possible area while orbiting around a planet or moon. It also has practical uses in fields like agriculture, where farmers need to determine the most efficient path for irrigation systems.

5. What are some real-world examples of maximizing area of trajectory?

Real-world examples of maximizing area of trajectory include long jump and javelin throw in athletics, golf shots, satellite orbits, missile trajectories, and even the path taken by a bee collecting nectar from multiple flowers. In addition, many amusement park rides and roller coasters are designed with the goal of maximizing the area of trajectory for a more exciting and thrilling experience.

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