A golfer imparts a speed of 30.3 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? (b) What is the longest “hole in one” that the golfer can make, if the ball does not roll when it hits the green? This is a question in my physics book in the chapter on kinematics in two dimensions.
The Correct Answer and Explanation is :
To solve this problem, we apply the principles of projectile motion.
Given:
- Initial speed of the ball, ( v_0 = 30.3 \, \text{m/s} )
- Launch angle for maximum range, ( \theta = 45^\circ ) (since ( \sin(2\theta) ) is maximized at ( 45^\circ ))
- Acceleration due to gravity, ( g = 9.8 \, \text{m/s}^2 )
(a) Time of flight
The time of flight for projectile motion is given by:
[
t_{\text{flight}} = \frac{2v_0 \sin\theta}{g}
]
Substituting values:
[
t_{\text{flight}} = \frac{2(30.3)(\sin 45^\circ)}{9.8}
]
Since ( \sin 45^\circ = \frac{\sqrt{2}}{2} ):
[
t_{\text{flight}} = \frac{2(30.3)(0.707)}{9.8} \approx \frac{42.88}{9.8} \approx 4.38 \, \text{seconds}
]
(b) Maximum range
The range of a projectile is given by:
[
R = \frac{v_0^2 \sin(2\theta)}{g}
]
For ( \theta = 45^\circ ), ( \sin(2\theta) = \sin(90^\circ) = 1 ):
[
R = \frac{(30.3)^2(1)}{9.8}
]
[
R = \frac{918.09}{9.8} \approx 93.7 \, \text{meters}
]
Explanation (300 words)
This problem involves understanding the motion of a projectile under the influence of gravity. A projectile launched at an angle experiences both horizontal and vertical components of motion. The horizontal velocity remains constant, while the vertical motion is influenced by gravity, causing the projectile to follow a parabolic trajectory.
For maximum range, the launch angle is ( 45^\circ ), as this angle maximizes the horizontal displacement by balancing the vertical and horizontal components of velocity. The time of flight depends only on the vertical motion. Using ( t = \frac{2v_0 \sin\theta}{g} ), we calculate how long the ball remains in the air based on its initial vertical velocity and the time taken to rise and fall symmetrically.
The range formula ( R = \frac{v_0^2 \sin(2\theta)}{g} ) accounts for the horizontal displacement during the flight time. At ( 45^\circ ), ( \sin(2\theta) = 1 ), giving the maximum range.
For ( v_0 = 30.3 \, \text{m/s} ), the time of flight is ( 4.38 \, \text{seconds} ), and the maximum range is ( 93.7 \, \text{meters} ). These values show how velocity, angle, and gravity influence the projectile’s trajectory.