Question

Suppose that an object is thrown down towards the ground from the top of the 1100-ft tall Sears
Tower with an initial velocity of – 20 feet per second (the negative velocity indicates that the object
is moving towards the ground). Given that the acceleration due to gravity of the object is a constant –
32 feet per second per second, determine how far above the ground the object is exactly six seconds
after being thrown.

Answer 1

h = Hi + Vi t + (1/2) a t^2
Hi = 1100
Vi = - 20
a = -32
so
h = 1100 -20 t - 16 t^2
if t=6
h = 1100 -20*6 - 16*36

= 404 ft

Answer 2

all this from integrating acceleration
a = -32 ft/s^2
so
dv/dt = -32
v = Vi - 32 t

dh/dt = v = Vi - 32 t
h = Hi + Vi t - (1/2)(32)t^2