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
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.
To determine the time the ball spends in the air, we can use the equation of motion for projectile motion. The ball is launched with an initial speed of 32.9 m/s, and assuming no air resistance, the only force acting on the ball is gravity.
(a) To find the time the ball spends in the air, we can use the fact that the vertical displacement of the ball when it lands on the green is zero. We can use the formula:
y = (v0 * t) + (0.5 * a * t^2)
Since the vertical displacement is zero, the equation becomes:
0 = (v0 * t) + (0.5 * (-9.8) * t^2)
Simplifying:
0 = 32.9t - 4.9t^2
Rearranging to quadratic form:
4.9t^2 - 32.9t = 0
Factoring out t:
t(4.9t - 32.9) = 0
So either t = 0 or 4.9t - 32.9 = 0.
Since time cannot be zero in this case, we solve the second equation:
4.9t - 32.9 = 0
4.9t = 32.9
t = 32.9 / 4.9
t ≈ 6.73 seconds
Therefore, the ball spends approximately 6.73 seconds in the air.
(b) To determine the longest "hole in one" the golfer can make, we need to find the horizontal distance traveled by the ball. Since the horizontal velocity of the ball remains constant throughout its motion, we can use the formula:
x = v0 * t
where x is the horizontal distance, v0 is the initial velocity, and t is the time the ball spends in the air.
Using the values given:
x = 32.9 m/s * 6.73 s
Calculating:
x ≈ 221.32 meters
Therefore, the longest "hole in one" the golfer can make is approximately 221.32 meters.