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

A box is dropped from a spacecraft moving horizontally at 27.0 m/s at a distance of 155 m above the surface of a moon. The rate of freefall acceleration on this airless moon is 2.79 m/s2. (a) How long does it take for the box to reach the moon's surface? (b) What is its horizontal displacement during this time? (c) What is its vertical velocity when it strikes the surface? (d) At what speed does the box strike the moon?

- physics -
**Damon**, Wednesday, December 15, 2010 at 5:47pm
initial 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 6:43pm
i dont understand. can u pls explain the solution in more detail?

- physics -
**Damon**, Wednesday, December 15, 2010 at 6:57pm
The 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 7:00pm
thank you so much Damon!

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