A golfer imparts a speed of 25.8 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?
max distance is when θ=45°
so, use that to figure x- and y-components of initial velocity,
To find out how much time the ball spends in the air, we can use the fact that the initial vertical velocity is zero (since the tee and the green are at the same elevation) and the acceleration due to gravity is -9.8 m/s².
(a) How much time does the ball spend in the air?
We can use the following kinematic equation to find the time of flight (t):
h = v₀t + (1/2)gt²
Where:
h = initial vertical displacement (in this case, zero)
v₀ = initial vertical velocity (in this case, zero)
g = acceleration due to gravity (-9.8 m/s²)
t = time of flight (what we want to find)
Plugging in the values and solving for t:
0 = 0 + (1/2)(-9.8)t²
0 = -4.9t²
t² = 0
t = 0 (since the only solution to t² = 0 is t = 0)
Therefore, the ball spends zero seconds (or no time) in the air. This implies that the ball doesn't go up or down but instead travels along a straight-line path parallel to the ground.
(b) What is the longest "hole in one" that the golfer can make, if the ball does not roll when it hits the green?
Since the ball spends no time in the air, its total horizontal distance traveled will be determined by its initial horizontal velocity (25.8 m/s) and the total time it takes to reach the ground (which is zero as calculated in part a).
d = v₀ * t
Where:
d = horizontal distance
v₀ = initial horizontal velocity
t = time of flight (zero in this case)
Plugging in the values:
d = 25.8 * 0
d = 0
Therefore, the longest "hole in one" that the golfer can make, assuming the ball does not roll when it hits the green, is a distance of zero meters.