Assuming the result of problem 5 for the range of a projectile, R = (2v^2/g) sin

Question

Assuming the result of problem 5 for the range of a projectile, R = (2v^2/g) sin cos , show that the maximum range is for = 45.

This is problem 5: A cannon standing on a flat field fires a cannonball with a muzzle velocity v, at an angle above horizontal. The cannonball thus initially has velocity components vx = v cos and vy = v sin . (a) Show that the cannon’s range (horizontal distance to where the cannonball falls) is given by the equation R = (2v^2/g) sin cos . (b) Interpret your equation in the cases of = 0 and = 90

Explanation / Answer

R = (2v^2/g) sin cos ( assume this equation)

R(45 degree) = (2v^2/g) sin 45 cos 45 = v^2/g ( sin(45)= cos(45) = 1/(2)1/2 , => sin 45 cos 45 = 1/2)

R(0 degree) = 0 as sin(0) = 0

R( 90 degree) = 0 as cos(0) = 0

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