From d = V^2sin(2µ)/g where d = the horizontal distance traveled in meters(ignoring air friction), V = the initial velocity in m/s, µ = the angle of the velocity to the horizontal and g = the acceleration due to gravity = 9.8m/sec^2, it is clear that the maximum distance traveled will occur when the launch angle is 45º and 2µ = 90º.
The maximum height reached derives from h = Vv^2(sin^2(µ))/2g.
Having the height reached, the air time derives from Vvf = Vvo - gt where Vvf = the final vertical component of the initial velocity = 0, Vvo = the initial vertical component of the velocity or t = VVvo/g.
A golfer imparts a speed of 32.9 m/s to a ball, and it travels the maximum possible distance before landing on the green. The tee and the green are at the same elevation.
(a) How much time does the ball spend in the air?
s
(b) What is the longest "hole in one" that the golfer can make, if the ball does not roll when it hits the green?
m
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