MMH 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?
The correct answer and explanation is:
To solve this, we use the physics of projectile motion. The golfer hits the ball at the optimal angle for maximum range, which is 45 degrees, and the tee and green are at the same elevation. We are given:
- Initial speed v=30.3 m/sv = 30.3 \, \text{m/s}
- Acceleration due to gravity g=9.8 m/s2g = 9.8 \, \text{m/s}^2
(a) Time of flight
The formula for total time of flight for a projectile launched at angle θ=45∘\theta = 45^\circ is: t=2vsinθgt = \frac{2v\sin\theta}{g}
Substituting the values: t=2⋅30.3⋅sin(45∘)9.8t = \frac{2 \cdot 30.3 \cdot \sin(45^\circ)}{9.8}
Since sin(45∘)=22≈0.7071\sin(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.7071, t=2⋅30.3⋅0.70719.8=42.8259.8≈4.37 secondst = \frac{2 \cdot 30.3 \cdot 0.7071}{9.8} = \frac{42.825}{9.8} \approx 4.37 \, \text{seconds}
Answer for (a):
Time in the air ≈ 4.37 seconds
(b) Maximum horizontal range (longest hole in one)
The formula for range RR of a projectile on level ground is: R=v2sin(2θ)gR = \frac{v^2 \sin(2\theta)}{g}
At 45°, sin(2θ)=sin(90∘)=1\sin(2\theta) = \sin(90^\circ) = 1, so: R=(30.3)29.8=918.099.8≈93.68 metersR = \frac{(30.3)^2}{9.8} = \frac{918.09}{9.8} \approx 93.68 \, \text{meters}
Answer for (b):
Longest hole in one ≈ 93.68 meters
Explanation (300 words)
This problem involves projectile motion, where an object is launched into the air and follows a curved path under the influence of gravity. For maximum horizontal distance, a projectile must be launched at an angle of 45 degrees, assuming that the starting and ending heights are the same. The golfer hits the ball at a speed of 30.3 m/s, and since both the tee and the green are at the same elevation, we can use the standard formulas for projectile motion.
First, to calculate how long the ball is in the air, we apply the time of flight formula, which is t=2vsinθgt = \frac{2v\sin\theta}{g}. At 45 degrees, the sine function gives a value of about 0.7071. Plugging in the numbers, we find the ball stays in the air for approximately 4.37 seconds.
Next, for the distance the ball travels—known as the range—we use the formula R=v2sin(2θ)gR = \frac{v^2 \sin(2\theta)}{g}. With a 45-degree launch angle, sin(2θ)\sin(2\theta) equals 1, maximizing the range. Substituting the given speed into the formula yields a maximum range of approximately 93.68 meters.
These calculations are based on ideal conditions, meaning there is no air resistance and the ground is flat. In real-life golfing situations, terrain and wind would affect the outcome, but in this idealized case, the answers represent the theoretical maximum time aloft and maximum distance a golf ball can travel under the given conditions.