Thursday

October 2, 2014

October 2, 2014

Posted by **Anonymous** on Thursday, September 5, 2013 at 5:21pm.

Part A: What is the magnitude of the object’s acceleration after it is released (relative to the ground)?

Part B: How long does it take to hit the ground?

- physics -
**Damon**, Thursday, September 5, 2013 at 5:56pmF = m a

-m g is the only force on the object

-m g = m a

a = -g = -9.81 m/s^2

now where is the balloon?

v = 0 + 3.3 t

15 = 3.3 t

t = 4.55 seconds to reach v = 15

h = (1/2)a t^2 = (1/2)(3.3)(4.55^2) = 34.2 m high

so we release the rock at 34.2 meters while moving up at 15 m/s

a = -9.81

h = 34.2 + 15 t - 4.9 t^2

when is h = 0 ? (hits ground)

4.9 t^2 -15 t - 34.2 = 0

t = [ 15 +/ - sqrt (225 + 670) ] /9.8

t = [ 15 + 29.9 ]/9.8

t = 4.58 seconds

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