Question #1224

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?

Expert's answer

For the vertical velocity projection the equations of the motion are:

v = at

h = at2/2;

v= 2.79 t

155 = 2.79t^{2}/2.

Thus

a) t = 10.54 s.

b) Sh = 27 m/s * 10.54s = 284.58 m

c) Vv = 2.79 * 10.54 = 29.4 m/s

d) V = sqrt (Vh^{2} + Vv^{2}) = sqrt (27^{2} + 29.4^{2}) = 39.9 m/s

v = at

h = at2/2;

v= 2.79 t

155 = 2.79t

Thus

a) t = 10.54 s.

b) Sh = 27 m/s * 10.54s = 284.58 m

c) Vv = 2.79 * 10.54 = 29.4 m/s

d) V = sqrt (Vh

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