A box is dropped from a spacecraft moving horizontally at 27.0 m/s at a distance

Question

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/s^2, How long does it take for the box to reach the moon's surface? What is its horizontal displacement during this time? What is its vertical velocity when it strikes the surface? At what speed does the box strike the moon? (0/2 submissions used) s (0/2 submissions used) m (0/2 submissions used) m/s (0/2 submissions used) m/s

Explanation / Answer

a)

Along the vertical direction

Voy = initial velocity = 0 m/s

a = acceleration = 2.79 m/s2

t = time taken

Y = vertical displacement = 155 m

using the equation

Y = Voy t + (0.5) a t2

155 = 0 t + (0.5) (2.79) t2

t = 10.54 sec

b)

Along the horizontal direction

Vox = initial velocity = 27 m/s

X = Vox t = 27 x 10.54 = 284.6 m

c)

Using

Vfy = Voy + at

Vfy = 0 + 2.79 x 10.54

Vfy = 29.41 m/s

d)

Speed = sqrt(Vfx2 + Vfy2) = sqrt(272 + 29.412) = 39.92 m/s

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