physics math question?

Question

a stone is dropped from a roof. a second stone is dropped 1.5 seconds later. how far apart are the stones when the second stone reaches a speed of 12 m/s

Answer 1

The first stone has
y(t)=-.5*g*t^2
the second stone has
y(t)=-.5*g*(t-1.5)^2

The speed of the second stone has
v(t)=g*(t-1.5)
when v(t)=12 m/s, solve for t
I will use g=10
12=10*t-15
2.7=t

so the first stone is at
y1=-.5*g*2.7^2
and the second stone is at
y2=-.5*g*(2.7-1.5)^2

Iy1-y2I=50*(2.7^2-1.2^2)

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