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?

To find the answers to the given questions, let's solve them step-by-step:

Step 1: Calculate the time the ball spends in the air.
To calculate the time the ball spends in the air, we'll use the following kinematic equation:

Δy = v₀t + (1/2)gt²

Since the tee and the green are at the same elevation, the change in height (Δy) is zero.

0 = v₀t + (1/2)gt²

Rearranging the equation, we get:

t = -v₀/g

Where:
t is the time spent in the air.
v₀ is the initial velocity of the ball.
g is the acceleration due to gravity (approximately 9.8 m/s²).

The given initial velocity is v₀ = 25.8 m/s.
Plugging in the values, we can calculate the time:

t = -25.8 / 9.8
t ≈ -2.63 s

Since time can't be negative, we take the positive value:

t ≈ 2.63 s

So, the ball spends approximately 2.63 seconds in the air.

Step 2: Calculate the longest "hole in one."
To calculate the longest "hole in one," we need to find the horizontal distance traveled by the ball.

Horizontal distance (d) can be calculated using the formula:

d = v₀ × t

Plugging in the values, we have:

d = 25.8 × 2.63
d ≈ 67.83 m

Therefore, the longest "hole in one" the golfer can make, assuming the ball doesn't roll when it hits the green, is approximately 67.83 meters.

To find the answers to the given questions, we need to consider the motion of the ball and use some basic principles of physics. Let's break down the problem step by step:

(a) How much time does the ball spend in the air?

To determine the time of flight, we can apply the equation of motion for an object under constant acceleration in the vertical direction (assuming no air resistance):

h = (1/2)gt^2

Where:
h is the maximum height reached by the ball (which we know is zero since the tee and green are at the same elevation)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time of flight

Since the maximum height is zero, we can rewrite the equation as:

0 = (1/2)gt^2

Simplifying, we get:

0 = gt^2

Since g is nonzero, the only solution for this equation is t = 0, which means the ball spends no time in the air. However, this is not a realistic result since we know the ball is supposed to be in the air. This is because we assume the ground is completely flat, and the ball does not experience any gravitational acceleration in the horizontal direction.

(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 distance covered by a projectile in horizontal motion can be determined using the equation:

d = v * t

Where:
d is the distance covered by the ball
v is the initial horizontal velocity (speed)
t is the time of flight (which we already know is zero)

Since the time of flight is zero, the distance covered by the ball is also zero. This means that the longest "hole in one" the golfer can make, assuming the ball does not roll when it hits the green, is zero meters. In other words, the ball will not reach the green without any time in the air.

However, it's worth noting that in real-life situations, golf balls spin and roll when they hit the ground. So, in practice, a ball can still travel a considerable distance after landing on the green.