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

In summary, a rock is thrown vertically upward from the surface of the moon at a velocity of 24m/s and reaches a height of s=24t-0.8t^2 meters in t seconds. To find the rock's velocity and acceleration as functions of time, one must first find the first and second derivatives of s. To determine the time it took for the rock to reach its highest point, set the derivative equal to 0 or complete the square in the quadratic function. The height reached by the rock can be found by plugging the time from part (b) into the equation. To find when the rock reached half its maximum height, divide the height found in part (c) by 2 and solve for
  • #1
jordan6292
1
0
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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  • #2
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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