1) 5 s
The vertical position of the object is given by
[tex]y(t) = h - \frac{1}{2}gt^2 = 400 - 16 t^2[/tex]
where
h=400 ft represents the initial height
g = 32 ft/s^2 is the acceleration of gravity
t is the time
We want to find the time t at which the object reaches the ground, so the time t at which
y(t) = 0
By substituting this into the equation, we find
[tex]0 = 400 - 16t^2\\t=\sqrt{\frac{400}{16}}=5 s[/tex]
2) 160 ft/s
The object is released from rest, so the initial velocity is zero
u = 0
The final vertical velocity can be found by using
[tex]v^2 - u^2 = 2ah[/tex]
where
v is the final velocity
a = 32 ft/s^2 is the acceleration of gravity
h = 400 ft is the vertical distance covered
Solving for v, we find
[tex]v=\sqrt{u^2 +2ay}=\sqrt{2(32 ft/s)(400 ft)}=160 ft/s[/tex]
