Posted by olivia on Monday, October 1, 2012 at 6:40pm.
Easy way is to assume that total energy is constant (of course debris might have a lot of air drag but there is not much we can do about that.
mass m (will cancel everywhere in this problem)
(1/2) m v^2 + m g h = total energy
at start h = 0 and v = 75
so
(1/2)(75)^2 = total energy/m = 2813
in obsolete English units as here g = 32 ft/s^2
so at 12 ft
(1/2) v^2 = 32(12)
solve for v
and at 20 ft
(1/2) v^2 = 32 (20) solve for v
note that the speed is the same upwards and downwards at the same height, just the direction changes.
at top
(1/2) v^2 = 0
so
32 h = 2813
at 12 feet
(1/2)v^2 + 32 (12) = 2813
at 20 feet
(1/2) v^2 + 32 (20) = 2813
a)
Height(t) = -16t^2 + 75t
Height ' (t) = -32t + 75 = 0 for a max of Height
-32t = -75
t = 75/32 seconds or appr 2.34 seconds
b) Height(75/32) = -16(75/32)^2 + 75(75/32) = 87.89 ft
c) when Height = 12
12 = -16t^2+ 75t
16t^2 - 75t + 12 = 0
t = 4.5216 sec (on the way down) or
t = .1659 seconds (on the way up)
velocity(t) = height ' (t)
= -32t + 75 , so we want it on the way down
velocity(4.5216) = -32(4.5216)+75 = -69.69 ft/sec
d)
same method as in c)
16t^2 - 75t + 20 = 0
t = 4.40 seconds (on the way down), or
t = .284 secpmds (on the way up
velocity(.284) = -32(.284)+75 = 65.92 ft/sec