A golfer gives a ball a maximum initial speed of 51.4 m/s.
(a) What is the longest possible hole-in-one for this golfer? Neglect any distance the ball might roll on the green and assume that the tee and the green are at the same level.
(b) What is the minimum speed of the ball during this hole-in-one shot?
The Correct Answer and Explanation is :
To determine the longest possible hole-in-one and the minimum speed of the golf ball during its flight, we can analyze the projectile motion of the ball.
(a) Longest Possible Hole-in-One
The range ( R ) of a projectile launched with an initial speed ( v_0 ) at an angle ( \theta ) is given by:
[
R = \frac{v_0^2 \sin(2\theta)}{g}
]
where ( g ) is the acceleration due to gravity (approximately ( 9.81 \, \text{m/s}^2 )).
To achieve the maximum range, the launch angle ( \theta ) should be ( 45^\circ ), because ( \sin(90^\circ) = 1 ), which maximizes the sine function.
Substituting ( v_0 = 51.4 \, \text{m/s} ), ( \theta = 45^\circ ), and ( g = 9.81 \, \text{m/s}^2 ) into the equation:
[
R = \frac{(51.4)^2 \times \sin(90^\circ)}{9.81} = \frac{2642.96}{9.81} \approx 269.31 \, \text{meters}
]
Therefore, the longest possible hole-in-one for this golfer, neglecting any roll on the green and assuming the tee and green are at the same level, is approximately 269.31 meters.
(b) Minimum Speed During the Hole-in-One Shot
During projectile motion, the speed of the ball varies, reaching its minimum at the peak of its trajectory. At this highest point, the vertical component of the velocity becomes zero, leaving only the horizontal component.
The horizontal component ( v_{0x} ) of the initial velocity is calculated as:
[
v_{0x} = v_0 \cos(\theta)
]
For ( \theta = 45^\circ ):
[
v_{0x} = 51.4 \times \cos(45^\circ) = 51.4 \times \frac{\sqrt{2}}{2} \approx 36.35 \, \text{m/s}
]
Thus, the minimum speed of the ball during its flight is approximately 36.35 meters per second, occurring at the peak of its trajectory.
Explanation
Projectile motion describes the behavior of an object launched into the air, subject only to the acceleration due to gravity. The trajectory is parabolic, with the object reaching its maximum height at the midpoint of its horizontal range.
The range equation ( R = \frac{v_0^2 \sin(2\theta)}{g} ) derives from the horizontal and vertical components of the motion. The horizontal distance depends on the initial speed, the angle of launch, and the time the projectile remains in the air. By setting the launch angle to ( 45^\circ ), we optimize the product ( \sin(2\theta) ), achieving the maximum possible range for a given initial speed.
At the peak of the trajectory, the vertical component of the velocity is zero because gravity has decelerated the upward motion to a stop before accelerating it downward. At this point, the speed of the projectile equals its horizontal component, which remains constant throughout the flight (assuming no air resistance). This horizontal speed represents the minimum speed during the projectile’s flight.
Understanding these principles allows golfers and engineers alike to predict and optimize the distances achievable under ideal conditions.