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February 1, 2015

Posted by **kevin giordano** on Wednesday, December 15, 2010 at 5:38pm.

- physics -
**Damon**, Wednesday, December 15, 2010 at 5:47pminitial speed vertical is zero so pure falling problem in the vertical plane

-155 = 0 + 0 t + (1/2)(-2.79) t^2

then horizontal distance = 27 t

solve for t

v = Vi + a t

v = 0 -2.79 t

total speed = sqrt ( v^2 + 27^2)

- physics -
**kevin giordano**, Wednesday, December 15, 2010 at 6:43pmi dont understand. can u pls explain the solution in more detail?

- physics -
**Damon**, Wednesday, December 15, 2010 at 6:57pmThe vertical problem is dropping a mass from a height of 155 meters. The horizontal speed has nothing to do with the vertical problem. The vertical component of velocity is zero at the beginning.

So how long to drop:

y = Xo + Vi t + (1/2) a t^2

here

x = -155 (or 155 down)

Xo = 0, starts at height we define as 0

Vi = 0 , zero speed down at start

a = - 2.79 (moon gravity down)

so

-155 = 0 + 0 t +(1/2)(-2.79) t^2

t^2 = 310/2.79

t = 10.5 seconds to fall 155 meters

For that time it is moving horizontaly at 27 m/s

so

distance = 27 * 10.5 = 285 meters

the vertical component is a*t

which is -2.79*10.5 = -29.3 at surface

so total speed = sqrt[(-29.3)^2 + 27^2]

- physics -
**kevin giordano**, Wednesday, December 15, 2010 at 7:00pmthank you so much Damon!

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