Answer: the frequency is every 27.322 days
I'd answer that but I can't text graphs and tables...
Answer:
about 602 milliseconds
Explanation:
The motion can be approximated by the equation ...
y = -4.9t^2 -22.8t +15.5
where t is the time since the arrow was released, and y is the distance above the ground.
When y=0, the arrow has hit the ground.
Using the quadratic formula, we find ...
t = (-(-22.8) ± √((-22.8)^2 -4(-4.9)(15.5)))/(2(-4.9))
= (22.8 ± √823.64)/(-9.8)
The positive solution is ...
t ≈ 0.60195193
It takes about 602 milliseconds for the arrow to reach the ground.
V=d/t
V=?
d=400m(4)
=1600m
t=6 min.
=360 s
V=1600m/360s
V=4.4m/s