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:
![y(t) = h - \frac{1}{2}gt^2 = 450 - \frac{1}{2}(9.8)t^2 [m] = 450 - 4.9 t^2 [m]](https://tex.z-dn.net/?f=y%28t%29%20%3D%20h%20-%20%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%20%3D%20450%20-%20%5Cfrac%7B1%7D%7B2%7D%289.8%29t%5E2%20%5Bm%5D%20%3D%20450%20-%204.9%20t%5E2%20%5Bm%5D%20%20)
b) The time the stone takes to reach the ground is the time t at which its vertical position y(t) becomes zero:

and if we solve it, we find

c) Since it is a uniform accelerated motion, the velocity of the stone at time t is given by

where the initial velocity is zero:

. The stone hits the ground at t=9.6 s, so its velocity at that time is

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
![y(t) = h -v_0 t - \frac{1}{2}gt^2 = 450 - 5t -4.9t^2 [m]](https://tex.z-dn.net/?f=y%28t%29%20%3D%20h%20-v_0%20t%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%20%3D%20450%20-%205t%20-4.9t%5E2%20%5Bm%5D%20)
where

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:

and by solving this equation, we find