You drop a 2 kg rock from the top of a 10 m tall wall,what is the velocity of the rock just before it hits the

you drop a 2 kg rock from the top of a 10 m tall wall,what is the velocity of the rock just before it hits the ground? assume g =9.8m/s2

As the rock drops, its vertical velocity increases 9.8 m/s each second.
Final velocity^2 = Initial velocity^2 + (2 * acceleration * distance)
Initial velocity = 0 m/s
Final velocity^2 = 2 * 9.8 * 10
Final velocity = (2 * 9.8 * 10)^0.5 = 14 m/s

OR

Distance = vi * t + ½ * g * t^2
vi = 0
10 = ½ * 9.8 * t^2
Time of fall = √(10 ÷ 4.9)

Final velocity = Initial velocity + g * t
Final velocity = 9.8 * √(10 ÷ 4.9) = 14 m/s

you can use either

vf^2 = v0^2 + 2 a d

vf=final velocity
v0=initial velocty =0
a = accel = 9.8m/s/s
d=distance = 10m

which gives

vf^2 = 0 + 2 g d

vf= Sqrt[2 g d]

or energy conservation

m g h = 1/2 m v^2 or v = Sqrt[2 gh]

to get v = Sqrt[2 x 9.8m/s/s x 10m] = 14m/s

Study kinematics: the question is trivial.