Applied Calculus
posted by Marie .
Suppose that an object is thrown down towards the ground from the top of the 1100ft 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.

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 
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
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