2c. We assume that the equations for ballistic motion in a vacuum apply, and the rate of downward acceleration is 32 ft/s². Then it will take ...
... (288 ft/s)/(32 ft/s²) = 9 s
for the rocket to reach its maximum height (its vertical velocity to decrease to zero).
2d. The trip down takes the same amount of time as the trip up, so the round trip time is ...
... (9 s)×2 = 18
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The acceleration due to gravity decreases the vertical velocity by the same amount (32 ft/s) each second. An equation for that might be ...
... v = 288 -32t
When the vertical velocity is zero, the rocket is no longer going up. It has reached its maximum height. Then the equation for that looks like ...
... 0 = 288 -32t
... 32t = 288
... t = 288/32 = 9 . . . . . compare to the solution above
After 9 seconds, the rocket reaches its maximum height.
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Simplifying the description of free-fall is a little harder. In physics terms, the potential energy the rocket has at its maximum height all gets converted to kinetic energy on the way down. At impact with the ground, the kinetic energy is the same as it was when the rocket was launched. That is, its velocity is 288 ft/s earthward at impact. The above equation for velocity still applies, but now we're solving for v = -288.
... -288 = 288 -32t
... -576 = -32t
... 576/32 = t = 18 . . . . . twice the time it took to reach maximum height.
The round trip takes 18 seconds.
