calculusA stone is dropped from the upper observation deck of a tower, 450 m above the ground. (Assume g = 9.8 m/s^2) (a) Find the distance (in meters) of the stone above ground level at time t....

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= 450-0.5*g*t^2

b) Distance (in meters) of the stone above ground level at the time t; S1= 450-0.5*g*t^2

When the stone reaches ground S1=0

Then 450-0.5*g*t^2 = 0

              0.5*g*t^2 = 450

                        t^2 = 450/(0.5g)

                        t^2 = 91.837

                             t= sqrt(91.837)

                             t= +9.58 OR -9.58

Since time is positive t= 9.58s

c)

If the velocity that the stone hits the round is V

dS/dt = V as stated earlier.

 V =dS/dt

    =d(0.5*g*t^2)/dt

    = 0.5*g*2t

At t=9.58s stone falls to the ground.

Then V= 0.5*9.8*9.58 = 46.94 m/s

d)

If the stone thows at 5m/s; then U= 5m/s as in (1)

S= U*t+0.5*a*t^2

S= 450m (tower height)

a = g (accelaration due to gravity)

Then;

450 = 5*t+0.5*9.8*t^2

    0= 0.5*9.8*t^2+5*t-450

So t1= (-5+sqrt(5^2-4*0.5*9.8*-450)/(2*0.5*9.58)

     t2 = (-5-sqrt(5^2-4*0.5*9.8*-450)/(2*0.5*9.58)

t1= 9.09

t2= -10.11

since the time is positive the time taken by the stone to reach ground is 9.09s