A tennis ball machine serves a ball vertically upward from a height of 2 feet with an initial speed of 110 feet per second. Neglecting air resistance and taking g ≈ 32 ft/s^2 downward, after how many seconds does the ball reach its maximum height?

Introduction / Context:
Vertical motion under gravity is a standard application of one dimensional kinematics. When a ball is thrown straight upward, its speed decreases due to the constant downward acceleration of gravity until it reaches maximum height where its instantaneous velocity becomes zero. This question asks you to determine the time taken for a tennis ball, served by a machine with a given initial speed, to reach its highest point. Understanding how to use basic equations of motion is essential for solving such problems.

Given Data / Assumptions:

• Initial height above ground h0 = 2 ft (this does not affect the time to reach the top, only the total height).
• Initial upward speed v0 = 110 ft/s.
• Acceleration due to gravity g ≈ 32 ft/s^2 downward.
• Air resistance is neglected, and motion is purely vertical.
• We need the time when the ball attains maximum height, that is, when its velocity becomes zero.

Concept / Approach:
For vertical motion with constant acceleration, we use the kinematic relation v = v0 + a * t, where v is velocity at time t, v0 is initial velocity, and a is acceleration. At maximum height, vertical velocity becomes zero, so we set v = 0 and solve for t. Here, initial velocity is upward and positive, while acceleration due to gravity is downward and taken as negative. The initial height does not enter this particular calculation because time to reach the top depends only on the change in velocity under constant acceleration.

Step-by-Step Solution:
Step 1: Choose upward as the positive direction. Then v0 = +110 ft/s and acceleration a = -32 ft/s^2. Step 2: At maximum height, the vertical velocity v becomes zero. Use the equation v = v0 + a * t. Step 3: Set v = 0 and substitute values: 0 = 110 + (-32) * t. Step 4: Rearrange to solve for t: -32 * t = -110, so t = 110 / 32. Step 5: Compute 110 / 32 = 3.4375 s, which rounds to approximately 3.44 s. Therefore, the ball reaches maximum height after about 3.44 seconds.

Verification / Alternative check:
We can cross check by using the symmetry of projectile motion. The time to rise to maximum height equals the time to fall back from maximum height to the same level if starting and ending speeds in magnitude are equal. The rise time is v0 / g in magnitude, which here is 110 / 32 ≈ 3.44 s. Another check is to estimate: if the speed were 96 ft/s, the time would be exactly 3 s (since 96 / 32 = 3). With a slightly higher speed of 110 ft/s, a rise time slightly above 3 s is expected, and 3.44 s fits this reasoning.

Why Other Options Are Wrong:
2.87 s: This corresponds to a smaller initial speed than given and does not satisfy the equation 0 = 110 - 32 * t. 4.65 s: This would imply v0 ≈ 149 ft/s if v0 = g * t, which is much larger than 110 ft/s, so it is inconsistent with the data. 5.41 s: This is even larger and would require an even higher initial speed to reach the maximum height after this time under the same acceleration.

Common Pitfalls:
One common mistake is to include the initial height in the time calculation, even though it influences only the maximum height, not the time to reach the top. Another error is to use the wrong sign for acceleration, which can lead to negative time or double counting. Some students also try to use the displacement equation unnecessarily, making the algebra more complex. To avoid these issues, remember that the simplest relation for time to maximum height is v = v0 + a * t with v set to zero.

Final Answer:
The tennis ball attains its maximum height after approximately 3.44 s from the moment it is served upward.