SOLUTION: at what angle of firing will a projectile's range be equal to its maximum height?

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Question 1188601: at what angle of firing will a projectile's range be equal to its maximum height?
Found 2 solutions by ikleyn, Alan3354:
Answer by ikleyn(53763) About Me  (Show Source):
You can put this solution on YOUR website!
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at what angle of firing will a projectile's range be equal to its maximum height?
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So, we consider the staring point and the ending point of the projectile at the same horizontal ground level.


Let "a" be the angle of firing (with the horizon) and let "v" be the magnitude of the initial speed.


Then v*sin(a) is the initial vertical speed and v*cos(a) is (unchangeable, constant) horizontal speed.


Vertical speed is the linear function of time from the ground level to the highest point,

THEREFORE, the average vertical speed during the flight up is  %28v%2Asin%28a%29%29%2F2.


The time moving up from the ground to the highest point is  t = %28v%2Asin%28a%29%29%2Fg,

THERFORE, the maximum height is the product of the average vertical speed  %28v%2Asin%28a%29%29%2F2

by the time to get the maximum height  %28v%2Asin%28a%29%29%2Fg

    h%5Bmax%5D = %28v%5E2%2Asin%5E2%28a%29%29%2F%282g%29.


The range is the maximum horizontal distance  r = v%2Acos%28a%29%2A2t = %282%2Av%5E2%2Acos%28a%29%2Asin%28a%29%29%2Fg


Notice that the time moving horizontally is twice the time to get the highest point ! (which is obvious . . . )


Our equation is the equality of h%5Bmax%5D and r,  or  h%5Bmax%5D = r,  which is

    %28v%5E2%2Asin%5E2%28a%29%29%2F%282g%29 = %282v%5E2%2Acos%28a%29%2Asin%28a%29%29%2Fg.


Canceling all common factors in the numerators and denominators, we get

    sin%28a%29%2F2 = 2*cos(a).


It implies  tan(a) = 4;  hence,  a = arctan(4) = 1.326 radians = 75.964 degrees.


ANSWER.  Under the imposed condition, the firing angle with the horizon is a = arctan(4) = 1.326 radians = 75.964 degrees.

Solved.



Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
at what angle of firing will a projectile's range be equal to its maximum height?
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Here's a different method:
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The path of the projectile is a parabola.
Make the parabola thru the points A(-1,0), B(0,2) and C(1,0)
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The parabola's equation is y = -2x^2 + 2
y' = -4x
@x = -1, y' = 4
4 is the slope of the parabola at x = -1, and is the tangent of the angle.
Angle = 75.964 degrees with the horizontal.
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This is called "high angle fire"