At the end of 4.0 s the velocity of the rock is 39.24 m/s and its displacement from the point where it was dropped is 78.48 m.
The rock dropped from the guard rail is a freely falling body and is accelerated downwards under acceleration due to gravity.
To calculate the velocity after t=4.0 s, use the equation of motion,
[tex]v=u+at[/tex]
Here, u is the rock's initial velocity, which is zero. It falls under acceleration due to gravity, hence a =g=9.81 m/s².
Therefore,
[tex]v=u+at\\ v=gt\\ =(9.81m/s^2)(4.0s)\\ =39.24m/s[/tex]
If it falls through a distance s, then, use the equation,
[tex]s=ut+\frac{1}2} gt^2[/tex]
Therefore,
[tex]s=ut+\frac{1}2} gt^2\\ s=\frac{1}2} gt^2\\ =\frac{1}2}(9.81m/s^2)(4.0s)^2\\ =78.48 m[/tex]
Thus, after 4.0 s, the rock has a downward velocity of 39.24 m/s and it would have traveled a distance of 78.48m.
