Question 1076665: Hi can someone help me with this problem please? THANK YOU!
The range of a projectile is modeled by the function r(theta)=1/16V^2sin(theta)cos(theta),where V is the initial velocity and theta is the angle at which the object is initially propelled. The maximum range is achieved when theta=45degrees. Use exact values to compute how many feet short of maximum the projectile falls if theta=67.5degrees and V=120ft/sec.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! (1/16)120^2*1/2; the 1/2 comes from multiplying sqrt (2)/2, which are the values of the sine and cosine at 45 degrees.
=450 feet.
for theta=67.5 degrees,
(1/16)120^2*0.3536, the last the product of the sin (67.5) and cos (67.5)
=318.24 feet, or 131.76 feet short of the maximum.
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