What angle will give the object the greatest range?

In summary, the angle at which a rifle should be pointed to shoot the maximum range is n which is equal to the vertical displacement of the rifle from the horizontal. The equation for x which is used to calculate the distance a projectile will travel is x=v_0 + 1/2(a)t^2.
  • #1
astroman707
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5

Homework Statement


1-A rifle can shoot a projectile with a velocity of 207m/s. At what angle should the rifle be pointed to give the maximum range?
2-Evaluate the maximum range

Homework Equations


N/A

The Attempt at a Solution


I considered at first that the problem seemed like it could be solved as an optimization problem. Since velocity-x is rcos(n), I used used x(t)=207cos(n)t to model the distance. I took the first derivative and got
v(t)=207cos(n)-tsin(n). I then realized that I couldn't set the derivative equal to zero and solve for t, because there was still (n) to deal with.
The book I'm using(3,000 solved problems in physics; schaum's) recommended to consider that the vertical displacement is zero, and solve for t in the equation x=v_0 + 1/2(a)t^2, giving t=x/v_0. Then it substituted that in for t in the same kinematic equation, but for the y-displacement. It then solved for x, used some trig identities, and then used a value provided in the book from a different problem(bad typo perhaps?).
Is there a much simpler way to solve this problem, or is the book's method the only way? Because it seems that the problem didn't give enough information, so it used another problem to help solve it; very confusing. It seems that calculus could be used for this, but I'm not sure.
 
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  • #2
astroman707 said:
x(t)=207cos(n)t
Looks good for the angle "n" with the horizontal.
astroman707 said:
x=v_0 + 1/2(a)t^2
That looks wrong (check the units -- the units of x and v_0 are different, correct?). That was from your book? It should be more like y(t)=vy_0t - 1/2(a)t^2, if y(0)=0.

The term vy_0 is related to the initial velocity and the sin(n), so write out the full equation and substitute it back into the equation for x(t). This will give you an equation for the time t versus the initial angle n. Then you can use a derivative to maximize x based on the angle n and the associated time in the air...
 
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  • #3
I’m fairly sure that the procedure you describe is the simplest way to do the problem. I think your description of the procedure is flawed in a couple of ways, but I’m sure that is just due to loose communication. I think if you study your book’s solution more carefully you will tidy up the details and then it won’t seem too complicated. Let us know if there is any part you don’t understand.
 
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  • #4
Thanks for your replies everyone! I’ve realized that the method the book suggests is the best solution, and I’ve used that method in a few other problems. I discovered that the issue was the fact that the book was requiring additional information from a previous problem, so finding the solution would have been impossible without the auxiliary problem.
 

1. What is the optimal angle for maximum range when launching an object?

The optimal angle for maximum range when launching an object is 45 degrees.

2. How does the launch angle affect the range of an object?

The launch angle determines the initial velocity and direction of the object, which ultimately affects the range. A launch angle of 45 degrees will result in the longest range.

3. Is there a mathematical equation to calculate the angle for maximum range?

Yes, the optimal angle for maximum range can be calculated using the equation: θ = tan^-1 (v^2/gR), where θ is the launch angle, v is the initial velocity, g is the acceleration due to gravity, and R is the range.

4. Can the optimal angle for maximum range vary depending on the object's mass and air resistance?

Yes, the optimal angle for maximum range can vary depending on the object's mass and air resistance. Objects with greater mass and more air resistance may require a slightly higher launch angle for maximum range.

5. Why is 45 degrees considered the optimal angle for maximum range?

45 degrees is considered the optimal angle for maximum range because it allows for the most balanced combination of horizontal and vertical velocity components, resulting in the longest distance traveled before hitting the ground.

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