Question

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

Answer 1

Answer:

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 + $\frac{1}{2}$ 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  =    $\frac{1}{2}$ at²    = $\frac{1}{2}$ x 9.8 x 3.8²   = 70.8m