1) 5 s
The vertical position of the object is given by
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
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
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