Since the question is based on calculus we need to know that

dS/dt = V

S= distance in meters

t= time in seconds

V= velocity in m/s

 

In general terms S= U*t+0.5*a*t^2

U = initial velosity

a= accelaration

 

a)

So if you put S= U*t+0.5*a*t^2 ------------- (1) when the stone releases from tower;

Since the stone falls in free fall U=0 and a=g

Then S=0.5*g*t^2 ---------------  (2)

The distance (in meters) of the stone above ground level will be

450-S

Therefore distance (in meters) of the stone above ground level at the time t; S1=...

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