A stone tumbles into a mine shaft strikes bottom after falling for 3.8 seconds. How deep is the mine shaft

70.8m

Explanation:

Given parameters:

Time of fall  = 3.8s

Unknown:

Depth of the mine  = ?

Solution:

To solve this problem, we use the appropriate motion equation as shown  below;

            S  = ut + at²  

S is the depth

u is the initial velocity  = 0m/s

t is the time taken

a is the acceleration due to gravity

  Now insert the parameters and solve;

    Since u  = 0;

    S  =     at²    = x 9.8 x 3.8²   = 70.8m

12000j is work done

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.