a) The stone moves along the vertical direction by unifom accelerated motion, with acceleration equal to g (gravitational acceleration), starting from initial position h above the ground and with initial velocity equal to zero. So, its vertical position follows the law:
[tex]y(t) = h - \frac{1}{2}gt^2 = 450 - \frac{1}{2}(9.8)t^2 [m] = 450 - 4.9 t^2 [m] [/tex]
b) The time the stone takes to reach the ground is the time t at which its vertical position y(t) becomes zero:
[tex]0=y(t) =450-4.9 t^2[/tex]
and if we solve it, we find
[tex]t= \sqrt{ \frac{450}{4.9} }=9.6 s [/tex]
c) Since it is a uniform accelerated motion, the velocity of the stone at time t is given by
[tex]v(t) = v_0 -gt=-gt[/tex]
where the initial velocity is zero: [tex]v_0 = 0[/tex]. The stone hits the ground at t=9.6 s, so its velocity at that time is
[tex]v(9.6s)=-gt=-(9.81 m/s)(9.6 s)=-94.2 m/s[/tex]
where the negative sign means it is directed downward.
d) In this case, since the initial velocity is not zero, the position at time t is given by
[tex]y(t) = h -v_0 t - \frac{1}{2}gt^2 = 450 - 5t -4.9t^2 [m] [/tex]
where [tex]v_0=5 m/s[/tex] is the initial velocity.
The time the stone takes to reach the ground is the time t such that y(t)=0, so we have:
[tex]450-5t-4.9t^2 =0[/tex]
and by solving this equation, we find
[tex]t=9.1 s[/tex]
