Answer:
(a) The distance ( s(t) ) of the stone above the ground at time ( t ) when dropped is given by:[ s(t) = 50 - \frac{1}{2} \times 9.8 \times t^2 ](b) To find the time it takes for the stone to reach the ground, set ( s(t) ) equal to 0 and solve for ( t ):[ 0 = 50 - \frac{1}{2} \times 9.8 \times t^2 ]Solve for ( t ) using the quadratic formula:[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]where ( a = -\frac{1}{2} \times 9.8 ), ( b = 0 ), and ( c = 50 ).[ t = \frac{-0 \pm \sqrt{0 - 4 \times -\frac{1}{2} \times 9.8 \times 50}}{2 \times -\frac{1}{2} \times 9.8} ]Simplify and solve for ( t ).(c) The velocity ( v(t) ) of the stone at time ( t ) is given by the derivative of ( s(t) ) with respect to time:[ v(t) = -gt ]For part (b), the time found in part (b) will be used to find the velocity.(d) If the stone is thrown downward with an initial velocity of 3 m/s, the equation for the distance ( s(t) ) becomes:[ s(t) = 50 + 3t - \frac{1}{2} \times 9.8 \times t^2 ]