Final answer:
The question pertains to the vertical motion of an object thrown upward, where kinematics is used to calculate the object's maximum height, velocity, and acceleration at various times, and the average velocity during the initial part of its ascent.
Step-by-step explanation:
The question asks about the motion of an object thrown vertically upward and involves using the principles of kinematics to find various parameters of its motion. We'll use the known acceleration due to gravity (approximately 32 ft/s2 or 9.8 m/s2 downward) and properties of projectile motion.
We're given that the object has a speed of 32 ft/sec at half its maximum height. This means at its maximum height, the vertical velocity will be 0 ft/sec, and it will have decelerated at a rate of 32 ft/s2. Using kinematic equations, we can calculate the max height (H) by using the fact that at half the height (H/2), it still has a vertical speed of 32 ft/s. We use the equation V2 = U2 + 2as, where V is the final velocity, U is the initial velocity, a is acceleration, and s is displacement. Rearranging, we get H=(32 ft/s)2/(2 * 32 ft/s2) = 16 ft, so the total height is 32 ft.
One second after being thrown upwards, the object will have slowed down due to the acceleration of gravity. It starts with an initial velocity, and after one second, it would have been slowed down by 32 ft/s (since gravity is -32 ft/s2). The exact initial velocity isn't given, so a numerical answer for velocity after 1 second can't be provided without that information. However, the acceleration remains constant at -32 ft/s2 throughout the motion unless otherwise stated. Three seconds after, similar principles apply: the object will have continued to decelerate (or accelerate downwards after reaching the peak).
The average velocity during the first half-second can be calculated using the initial velocity and the final velocity at the end of the half-second. Assuming constant acceleration, the average velocity is the arithmetic mean of the initial and final velocities.