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August 2, 2014

August 2, 2014

Posted by **Marie** on Friday, April 20, 2012 at 4:41pm.

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.

- Applied Calculus -
**Damon**, Friday, April 20, 2012 at 6:59pmh = 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

- Applied Calculus -
**Damon**, Friday, April 20, 2012 at 7:02pmall 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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