What is the Physics Behind a Rock Thrown on the Moon?

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The discussion focuses on the physics of a rock thrown vertically on the Moon, with a given initial velocity of 24 m/s. The height of the rock as a function of time is expressed as s = 24t - 0.8t^2 meters. Key calculations include finding the rock's velocity and acceleration, determining the time to reach the highest point, calculating the maximum height, and identifying when it reaches half that height. The suggestions provided emphasize using derivatives and solving quadratic equations to find the necessary values. The discussion highlights the unique aspects of motion under the Moon's gravity, which is less than that of Earth.
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a rock is thrown vertically upward fom the surface of the moon at a velocity of 24m/s(about 86km/h)reaches a height of s=24t-0.8t^2meters in t seconds.

(a)find the rocks velocity and acceleration as functions of time?(accleration of gravity on the moon)
(b)how long did it take the rock to reach its highest point?
(c)how high did the rock go?
(d)when did the rock reach half its maximum height?
(e)how long was the rock aloft?
 
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well, you can't expect us to solve the problem for you! Here are a few suggestions.

(a) Find the first and second derivatives of s of course.
(b) You can find that time by setting the derivative equal to 0 or, conversely, by completing the square in the quadratic function.
(c) Put the t from (b) in the equation.
(d) After finding the height in (c), divide by 2, put the s equal to that and solve for t. there will be two solutions, of course.
(e) Set the height equal to 0 and solve for t. Again there will be two solutions. It should be clear which is the one you want.
 

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