Numerical: Solve by Three Equations of Motion

A stone is dropped from a height of 40m.
(a) How much time will it take to reach the ground?
(b) With what velocity will it strike the ground?

A)
s=ut+(1/2)at^2
u= initial velocity=0
a=acceleration=9.8
t=time=?
s=distance=40
since u=0 ut=0
so s=(1/2)at^2
t^2=2s/a
t=sqrt(2s/a)
t=sqrt(2(40)/(9.8))
t=2.86 secs
B)
v=at
v=(9.8)(2.86)
v=28m/s

I can't see why you would use three equations, you only need two.

You chose "Best Answer" while I was typing mine, so I'll paste it here. I'm pretty horrified that something without units in the work could be considered an answer at all, never mind "best".

This is a constant acceleration problem. There are only two constant acceleration equations worth ...

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remembering, since they are related by calculus:

x_f = x_i + v_i t + ½ a t²

v_f = v_i + a t

all the other equations that people use can be derived from these two using algebra, and by the time you get to constant acceleration kinematics, I trust algebra isn't a problem.

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Choose a coordinate system. I'll put the origin at the drop point, with positive downward. Therefore:

x_f = +40 m
x_i = 0 m
v_i = 0 m/s
v_f = ?
a = +9.8 m/s²
t = ?

Substituting the zero values into the first equation yields

x_f = ½ a t²

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Solve for t

t = √[ 2 x_f / a ] = √[ 2 (+40 m) / (+9.8 m/s² ] = ±2.86 s

Reject the negative answer, as the stone cannot hit the ground before it is dropped.

t = +2.86 s

Now the second equation can be used to find the final velocity