+1904 An object is propelled upward at an angle, \(\Theta \), \(45°<\Theta <90°,\)to the horizontal with an initinal velocity of \(v_0\) feet per second from the base of a plane that makes an angle 45° with the horizontal. If air resistance is ignored, the distance \(R\) that travels up the inclined plane is given by the function \(R(\Theta) = \frac{v_0\sqrt{2}}{16}\times cos(\Theta)\times(sin(\Theta)-cos(\Theta))\)
(a) Show that \(R(\Theta) = \frac{v_0\sqrt{2}}{32}\times[sin(2\Theta)-cos(2\Theta)-1]\)
(b) In calculus, you will be asked to find the angle \(\Theta \) that maximizes \(R\) by solving the equation \(sin(2\Theta)+cos(2\Theta) = 0\). Solve this equation for \(\Theta.\)
(c) What is the maxium distance \(R\) if \(v_0 = 32\) feet per second?
(d) Graph \(R = R(\Theta),\) \(45°≤\Theta ≤90°,\) and find the angle \(\Theta \) that maximizes the distance. Use \(v_0 = 32\) feet per second. Compare the results with the answers found eariler
+23252 a) To show that v0 sqrt(2) / 32 x [sin(2A) - cos(2A) - 1] equals v0 sqrt(2) / 16 x cos(A) x [sin(A) - cos(A)]
---> v0 sqrt(2) / 16 x (1/2)[sin(2A) - cos(2A) - 1] equals v0 sqrt(2) / 16 x cos(A) x [sin(A) - cos(A)]
---> (1/2)[sin(2A) - cos(2A) - 1] equals cos(A) x [sin(A) - cos(A)]
---> But: sin(2A) = 2sin(A)cos(A) and cos(2A) = 2cos2(A) - 1
---> Left side:
(1/2)[2sin(A)cos(A) - (2cos2(A) - 1) - 1]
= (1/2)[2sin(A)cos(A) - 2cos2(A) + 1 - 1]
= (1/2)[2sin(A)cos(A) - 2cos2(A)]
= sin(A)cos(A) - cos2(A)
= cos(A)[sin(A) - cos(A) <--- which equals the right side.
b) sin(2A) + cos(2A ) = 0
---> sin(2A) = -cos(2A) (now divide both sides by cos(2A):
---> sin(2A) / cos(2A) = -1
---> tan(2A) = -1
---> 2A = 135°
---> A = 67.5°
c) Place these values into the equation, and you will find the maximum distance is 0.586 feet (approx).
