Answer:
15.49m/s
Explanation:
Using one of the equations of motion;
[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2gs
Where;
v = the final velocity of the rock.
=> This is the speed at which the rock strikes the water.
u = initial velocity of the rock.
=> The rock is just dropped from that height. That means the initial velocity is zero (0).
=> u = 0
g = the acceleration due to gravity.
=> Since the rock moves downwards, the acceleration due to gravity is positive and let it have a value of 10m/[tex]s^{2}[/tex]
=> g = 10m/[tex]s^{2}[/tex]
s = the distance covered by the rock
=> s = 12m
Substituting these values into the equation above gives
[tex]v^{2}[/tex] = [tex]0^{2}[/tex] + 2(10 x 12)
[tex]v^{2}[/tex] = 240
v = [tex]\sqrt{240}[/tex]
v = 15.49m/s
Therefore the velocity(speed) with the rock strikes the water is 15.49m/s
