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?

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 don't 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!

To answer these questions, we can use the equations of motion for freefall:

1. For (a) To find the time it takes for the box to reach the moon's surface, we can use the equation:
distance = initial velocity × time + (1/2) × acceleration × time^2

Plugging in the values:
Distance = 155 m
Initial velocity = 0 m/s (since the box is dropped, it has no initial velocity)
Acceleration = 2.79 m/s^2
Time = ?

The equation becomes:
155 m = 0 × time + (1/2) × 2.79 m/s^2 × time^2

Rearranging the equation:
(1/2) × 2.79 m/s^2 × time^2 = 155 m

Solving for time:
(1/2) × 2.79 m/s^2 × time^2 = 155 m
time^2 = (155 m × 2) / 2.79 m/s^2
time^2 = 222.22 s^2
time ≈ √(222.22 s^2) ≈ 14.92 s

So, it takes approximately 14.92 seconds for the box to reach the moon's surface.

2. For (b) To find the horizontal displacement during this time, we can use the equation:
displacement = initial velocity × time

Plugging in the values:
Initial velocity = 27.0 m/s
Time = 14.92 s
Displacement = ?

The equation becomes:
Displacement = 27.0 m/s × 14.92 s

Displacement ≈ 402.84 m

So, the horizontal displacement of the box during this time is approximately 402.84 meters.

3. For (c) To find the vertical velocity when it strikes the surface, we can use the equation:
final velocity = initial velocity + acceleration × time

Plugging in the values:
Initial velocity = 0 m/s
Acceleration = 2.79 m/s^2
Time = 14.92 s
Final velocity = ?

The equation becomes:
Final velocity = 0 m/s + 2.79 m/s^2 × 14.92 s

Final velocity ≈ 41.5 m/s

So, the vertical velocity of the box when it strikes the surface is approximately 41.5 m/s.

4. For (d) To find the speed at which the box strikes the moon, we need to calculate the net velocity of the box. We can use the Pythagorean theorem to find the magnitude of the velocity vector:

magnitude of velocity = √(horizontal velocity^2 + vertical velocity^2)

Plugging in the values:
Horizontal velocity = 27.0 m/s
Vertical velocity = -41.5 m/s (negative because it is moving downward)
Magnitude of velocity = ?

The equation becomes:
Magnitude of velocity = √(27.0 m/s)^2 + (-41.5 m/s)^2

Magnitude of velocity ≈ √(729 m^2/s^2 + 1712.25 m^2/s^2)
Magnitude of velocity ≈ √2441.25 m^2/s^2
Magnitude of velocity ≈ 49.41 m/s

So, the box strikes the moon with a speed of approximately 49.41 m/s.