Angle between 45 and 90 degrees: An object is propelled upward to the horizontal

Question

Angle between 45 and 90 degrees:



An object is propelled upward to the horizontal with an initial velocity of Vo feet per second from the base of an inclined plane that makes an angle of 45 degrees with the horizontal. If air resistance is ignored, the distance r that the object travels up the inclined plane is given by



R(x) = [v^2 * square roote of 2]/32 * [sin (2x) = cos (2x) - 1]


I know how to put into my calculator an get a mximum distance. But how do I figure out the maximum angle that provides the maximum distance?

Explanation / Answer

This is not a problem of physics, its a simple problem of maxima and minima and related to calculus basics. STEPS: 1. Differentiate the equation with respect to only variable "x". 2. Equate the result of (1) as zero. 3. Again differentiate the result of (1). 4. check that after double differentiation, your result is negative or not. a) If its negative, it will give you maxima. b) If its positive, it will give you minima. For your case here is the solution. *your prob is not clear, how come equality sign in the middle of an expression?? 1. For distance to be maximum, denominator of the expression must be minimum. 2. Apply above 4 procedures. 3. You will get the answer accordingly.

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