During what intervals of the domain of the water balloons height increasing

Answer 1

a water balloon, like most objects that have greater density than air,

falls at the rate of approximately-32 feet per second per second

or -9.8 meters per second per second

the equation for the height is h(t) = -16t^2 + vot + ho

where vo = initial vertical velocity of the water balloon

ho = iniitial height of the water balloon

it reaches maximum height when h'(t) = -32t + vo =0

when 32t = vo

when t = vo/32 seconds

its maximum height is h(vo/32) = -16(vo/32)^2 + vo(vo/32) + ho

It is increasing in height from time t=0 until time t= vo/32

the interval of the domain when h(t) is increasing is (0, vo/32)

if initial speed vo =0, or <0, then the water balloon is never increasing in height.

if vo=32 feet per second upward, then the water balloon is increasing in height for one second, from t=0 to t=1

for the interval (0, 1). At 1 second, it reaches maximum height, after 1 second it fall back down, decreasing in height. Maximum height = initial height ho + 16 feet.

if initial velocity = vo = 64 ft/sec, then the interval is (0,2), increasing in height for 2 seconds

This is all premised on denying Galileo's

proof that the heavy object falls exactly as fast as the lighter object.

as well as ignoring air resistence & buoyancy. But to simplify things, and that's how science works, getting a simplified model. As Prof. Leonard said in a seminar in the Big Bang, with his opening joke, consider a spherical cow. Every simplification ignores some reality. But it's necessary to focus on the most important variables. h(t) = -16t^2 + vot + ho is an over simplification, but it focuses on what most matters in calculating height from time