A rock falls from a cliff and travels downward for 6 seconds before striking the ground. How fast was the rock moving when it hits the ground? Assume there is no air resistance.

Given parameters:

duration of fall  = 6s

Unknown:

Speed of the rock before it hits the ground  = ?

Solution:

To solve this problem, we apply one of the appropriate motion equations;

           V = u + gt

Where V is the final velocity

           u is the initial velocity

           t is the time taken

           g is the acceleration due to gravity  = 9.8m/s²

The initial velocity is 0;

     So;

              V = gt

Input the parameters and solve for V;

                V = 9.8 x 6  = 58.8m/s

The velocity before hitting the ground is 58.8m/s

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"