Posted by kevin giordano on Wednesday, December 15, 2010 at 5:38pm.
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)
i dont understand. can u pls explain the solution in more detail?
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]
thank you so much Damon!
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