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Verified Answer :

syed514

, let us use the expression v = ./2gS in both the cases. 
a) In the first case g = 9.8 m/s^2 and S = 12 m. 
So v = 14.21 m/s 

stigawithfun

Answer:

15.49m/s

Explanation:

Using one of the equations of motion;

\(v^{2}\) = \(u^{2}\) + 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/\(s^{2}\)

=> g = 10m/\(s^{2}\)

s = the distance covered by the rock

=> s = 12m

Substituting these values into the equation above gives

\(v^{2}\) = \(0^{2}\) + 2(10 x 12)

\(v^{2}\) = 240

v = \(\sqrt{240}\)

v = 15.49m/s

Therefore the velocity(speed) with the rock strikes the water is 15.49m/s

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