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How do you differentiate 2sinAcosA?
First using the double angle formula; 2sinAcosA = sin(2A) Therefore d/dA (sin2A) = cos(2A) * 2 = 2cos(2A) We multiply by the two due to the chain rule, where we must multiply by the derivative of the angle.
What is sin-cos?
Sin is sine. Cos is cosine. http://en.wikipedia.org/wiki/Sine_curve http://en.wikipedia.org/wiki/Cosine_curve In terms of trigonometric identities sin2A=2sinAcosA cos2A=cos2A-sin2A sin2A-cos2A=2sinAcosA-cos2A+sin2A === === sin(A) - cos(A) = sqrt(2)sin(A-45)
How do you proof the formula sin2A equals 2sinAcosA?
First, note that sin(a+b)=sin(a)cos(b)+sin(b)cos(a)[For a proof, see: www.mathsroom.co.uk/downloads/Compound_Angle_Proof.pptFor the case of b=a, we have:sin (a+a)=sin(a)cos(a)+sin(a)cos(a)sin (2a)=2*sin(a)cos(a)
How can you determine the landing point of a horizontally launched projectile if you know the distance it will fall and the initial velocity?
Firing from a hill Firing a projectile from an elevated position increases its range. If you know the initial velocity, you should be able to use the usual formulas for displacement (distance) in the horizontal and vertical directions to determine the initial vertical position.1 When you say the initial velocity is known, I assume that includes magnitude and direction. Since velocity is a vector, you should be able to calculate the vertical and horizontal components.2 If you know the horizontal velocity and the horizontal displacement (distance traveled), you should be able to calculate the time in flight. Once you determine the time in flight, you should be able to use that value in the formula for vertical displacement to determine the initial vertical displacement. Hint: The vertical displacement of the projectile when it hits the ground is zero (assuming you have selected the origin -- the axes of the plane in which the projectile is moving -- properly). ----------- 1. d = d0 + V0t + [1/2]at2, where d0 is the initial displacement, v0 is the initial velocity, and a is acceleration. For motion in the vertical direction, a = -g. For motion in the horizontal direction, a = 0 (for projectile problems). 2. Vx = Vcos(theta); Vy = Vsin(theta), where theta is the angle of elevation. Maximum range is achieved when theta = 45 degrees. At that angle, Vx = Vy.*********PLEASE NOTE: formatting has been messed up in this so things that are supposed to be raised to a power, the number is not a superscript. This needs to be relooked at.Maximum range is achieved when theta=45o only if the vertical displacement is zero (i.e. the projectile begins and ends at the same elevation). If launched from a certain height h, the angle for maximum range is given byanglemax = 1/2 cos-1 [(gh)/(v^2 + gh)]Returning to the problem, let h = launch height, R = horizontal distance from base of launch site to landing spot, V = launch speed, A = launch angle and T = time in air. The horizontal component of the launch velocity is constant since there is no acceleration in that direction. Therefore:Vx = R/TV cosA = R/TSolving for T:T = R/[V cosA]Consider the vertical part of the problem. This solution is given for a projectile launched from an angle above the horizontal so that the initial vertical component of the velocity is positive when the acceleration due to gravity (g) is negative. Also assumed is that the launch position is above the landing position. Let the initial position be the origin.d = do + viT + 1/2aT^2-h = (V sinA)T - 1/2gT^2Substituting for T:(equation A) -h = [VR sinA ]/[V cosA] - [gR^2]/[2V^2 cos2A](equation B) h = -R [tanA] + [gR^2]/[2V^2cos2A]******If you want to find the launch angle for a given height and launch speed that gives the maximum range, multiply both sides of equation A by 2cos2(theta) and rearrange to get:(g/v^2)R^2 - (2sinAcosA)R - 2hcos2A=0Using these trig identities:2sinAcosA = sin2Acos2A = 1/2 [1 + cos2A]the equation becomes:(g/v^2)R^2 - (sin2A)R - h[1+cos2A] = 0Solving for R using the quadratic formula:R(A) = v^2/(2g)[sin2A + (sin22A + (8gh/v^2)cos2A)1/2]Find the derivative ofR(A): R'(A)=v^2/(2g)[2cos2A+1/2(sin22A+(8gh/v^2)cos2A)-1/2(4sin2Acos2A+(8gh/v^2)(-2sinAcosA))]Set this equal to zero to find angle (A) for maximum range (R).Rearrange and use some trig identities to get:1/(cos22A) - v^2/(gh)(1/cos2A) - (1+v^2/(gh))=0Use the quadratic formula to solve for 1/(cos2A):1/(cos2A) = (1/2)[v^2/(gh) + (v^4/(g^2h^2)+4(1+v^2/(gh)))1/2] 1/(cos2A) = v^2/(gh) + 1A = 1/2cos-1[(gh)/(v^2+gh)]
