Final answer:
To determine the time a projectile fired upward at an initial velocity of 96 feet per second returns to ground level, and when it exceeds a height of 128 feet, we use kinematic equations accounting for initial velocity and acceleration due to gravity.
Step-by-step explanation:
The question is related to classic physics problems involving projectile motion. To solve these problems, we use kinematic equations and principles of motion under the influence of gravity. In this case, the vertical motion of a projectile fired upward is considered. We will apply the following kinematic equation that relates displacement Δy, initial velocity vy_0, acceleration a, and time t:
Δy = vy_0 * t + 0.5 * a * t^2
For a projectile fired upwards at an initial velocity of 96 feet per second:
Part A: When the projectile returns to ground level, its displacement Δy is zero. Using the quadratic formula, we can find the time at which the projectile will be back at ground level.
Part B: To find when the height will exceed 128 feet, we set Δy > 128 feet and solve for the time interval using the same equation. Interval notation is used to represent the time period during which the projectile's height is greater than 128 feet.
Note that for these calculations:
- Gravity (g) is -32 feet per second squared, where the negative sign indicates that gravity acts downward.
- The initial vertical velocity (vy_0) is 96 feet per second upwards.
- For height exceeding 128 feet, we'll solve for the time when the projectile is ascending and descending passing the height of 128 feet.
By applying the necessary physics equations and conceptual understanding of projectile motion, we can determine the appropriate times for each part of the question.