Question

Free Fall on a Small Airless Planet A rock released from rest to fall on a small airless planet falls \(y = g t ^ { 2 } \mathrm { m }\) in \(t \mathrm { sec } , g\) a constant. Suppose that the rock falls to the bottom of a crevasse 20\(\mathrm { m }\) below and reaches the bottom in 4\(\mathrm { sec. }\) (a) Find the value of \(g .\) (b) Find the average speed for the fall. (c) With what speed did the rock hit the bottom?

Short answer

The constant 'g' is 1.25 \(m/sec^2\). The average speed is 5 \(m/sec\). The final speed is 5 \(m/sec\).

Step-by-step solution

Understanding and Formulating the Problem

The rock falls 20m deep for 4 seconds. The scenario is a usual free fall situation so we can use the second equation of motion to solve for \(g\), then find the average and final speed.

Solve for \(g\)

The second equation of motion is \(y = g t^2\). Substitute \(y = 20m \) and \(t = 4sec\) into the equation:\(20m = g (4sec)^2\)By solving this equation, we can derive the value of \(g\).

Solve for Average Speed.

The average speed can be calculated using the formula: \(Average Speed = Total distance ÷ Total time\)Substitute the given total distance and total time: \(Average Speed = 20m ÷ 4sec\).

Solve for Final Speed.

The third equation of motion is \(v = g t\). Again, substitute calculated \(g\) and \(t = 4 sec\) into the equation to compute for \(v\).