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?
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To find the answers to these questions, we can use the kinematic equations of motion. We will separate the motion of the box into its vertical and horizontal components.
(a) To find the time it takes for the box to reach the moon's surface, we will use the equation:
y = y0 + v0y * t + (1/2) * a * t^2
Where:
y = final vertical displacement (0 m, as the box reaches the surface)
y0 = initial vertical displacement (155 m)
v0y = initial vertical velocity (0 m/s, as the box is dropped)
a = acceleration (2.79 m/s^2)
t = time
Substituting the values into the equation:
0 = 155 + 0 * t + (1/2) * 2.79 * t^2
Simplifying the equation:
1.395t^2 = 155
Solving for t:
t^2 = 155 / 1.395
t^2 = 111.111
t ≈ 10.54 seconds
Therefore, it takes approximately 10.54 seconds for the box to reach the moon's surface.
(b) The horizontal displacement during this time can be found using the equation:
x = x0 + v0x * t
Where:
x = horizontal displacement
x0 = initial horizontal displacement (0 m, as the motion is horizontal)
v0x = initial horizontal velocity (27.0 m/s)
t = time (10.54 seconds)
Substituting the values into the equation:
x = 0 + 27.0 * 10.54
Calculating the result:
x ≈ 284.58 meters
Therefore, the horizontal displacement during this time is approximately 284.58 meters.
(c) The vertical velocity when the box strikes the surface can be found using the equation:
v = v0y + a * t
Where:
v = final vertical velocity
v0y = initial vertical velocity (0 m/s)
a = acceleration (2.79 m/s^2)
t = time (10.54 seconds)
Substituting the values into the equation:
v = 0 + 2.79 * 10.54
Calculating the result:
v ≈ 29.3546 m/s
Therefore, the vertical velocity when the box strikes the surface is approximately 29.3546 m/s.
(d) The speed at which the box strikes the moon can be found using the Pythagorean theorem:
speed = √(v^2x + v^2y)
Where:
v = final velocity
vx = horizontal velocity (27.0 m/s)
vy = vertical velocity (29.3546 m/s)
Substituting the values into the equation:
speed = √((27.0)^2 + (29.3546)^2)
Calculating the result:
speed ≈ 39.27 m/s
Therefore, the box strikes the moon at approximately 39.27 m/s.