Question

A stone is dropped from the upper observation deck (the space deck) of the cn tower, 450 m above the ground.

Answer 1

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:
$y(t) = h - \frac{1}{2}gt^2 = 450 - \frac{1}{2}(9.8)t^2 [m] = 450 - 4.9 t^2 [m]$

b) The time the stone takes to reach the ground is the time t at which its vertical position y(t) becomes zero:
$0=y(t) =450-4.9 t^2$
and if we solve it, we find
$t= \sqrt{ \frac{450}{4.9} }=9.6 s$

c) Since it is a uniform accelerated motion, the velocity of the stone at time t is given by
$v(t) = v_0 -gt=-gt$
where the initial velocity is zero: $v_0 = 0$. The stone hits the ground at t=9.6 s, so its velocity at that time is
$v(9.6s)=-gt=-(9.81 m/s)(9.6 s)=-94.2 m/s$
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
$y(t) = h -v_0 t - \frac{1}{2}gt^2 = 450 - 5t -4.9t^2 [m]$
where $v_0=5 m/s$ 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:
$450-5t-4.9t^2 =0$
and by solving this equation, we find
$t=9.1 s$