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

  • Thread starter Thread starter jordan6292
  • Start date Start date
  • Tags Tags
    Word problem
jordan6292
Messages
1
Reaction score
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?
 
Physics news on Phys.org
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top