Question

While attempting a landing on the moon, astronauts had to change their landing site and land at a spot that was 4 kilometers away from the original site. Assuming that they were at a height of 137 meters, calculate the horizontal velocity of the spacecraft during touchdown if it lands in a free-fall mode without using retro engines. Consider gravity = 1.63 meters/second^2.

A) 143.23 meters/second

B) 233.33 meters/second

C) 308.88 meters/second

D) 333.44 meters/second

Answer 1

308.8 m per s

Step-by-step explanation:

Answer 2

The closest answer would be C.
The choices given do not give the exact value.

To answer this, you just need to remember the main formula:


`d = Vit + (1)/(2)gt^(2)`

Where:
d = distance/displacement
g = acceleration due to gravity
t = time in flight
Vi = initial velocity.

With this formula, you derive all the formulas you need to look for certain components. You need to keep in mind of the following:

---if you are looking for a vertical component(y), you need to use values of vertical motion.


`dy = Viyt + (1)/(2)gt^(2)`

*Viy is always 0m/s at the beginning of a free-fall.


`dy = (0m/s)t + (1)/(2)gt^(2)`


`dy = (1)/(2)gt^(2)`

---If you are looking for a horizontal component(x), you need to use values of horizontal motion.


`dx = Vixt + (1)/(2)gt^(2)`

*g is always 0m/s² when taking horizontal motion into account.


`dx = Vixt + (1)/(2)(0m/s^(2))t^(2)`


`dx = Vixt`
--- time is the only value that is both vertical and horizontal.

Okay, let's get back to solving your problem. Let's see what your given is first:
dy = 137m (as long as it refers to height, it is vertical distance)
dx = 4km (the word, far or away usually indicates horizontal distance)
g = 1.63m/s²

The question is how fast was it going horizontally and we can derive it from our equation:


`dx = Vixt`

We use this because x means horizontal. But notice that we do not have time yet. So how are we going to solve this 2 variables missing? The key is that time is a horizontal and vertical component. Whatever time it took moving horizontally, it is the same vertically as well. So we use the vertical formula to derive time:


`dy = (1)/(2)gt^(2)`

`(2dy)/(g)=t^(2)`

`\sqrt{(2dy)/(g)} = \sqrt{t^(2)}`

`\sqrt{(2dy)/(g)} = t`

Now plug in what you know and solve for what you don't know:


`\sqrt{(2(137m))/(1.63m/s^(2))} = t`

`\sqrt{(274m)/(1.63m/s^(2))} = t`

`\sqrt{168.098s^(2)}=t`

`12.965s = t`

The total time in flight is 12.965s.
Let's round it off to 13s.

Now that we know that, we can use this in the horizontal formula:

`dx = Vixt`

`4km = Vix(13s)`

Hold up! Look at the unit of the horizontal distance. It is in km but all our units are expressed in m so we need to convert that first.

1km = 1,000m
4km = 4,000m

Our new horizontal distance is 4,000m.

Okay, let's wrap this up by solving for what is asked for, using all the derived values.

`dx = Vixt`

`4,000m = Vix(13s)`

`(4,000m)/(13s) = Vix`

`308m/s= Vix`

The horizontal velocity is 308m/s.

Such a long explanation I know, but hopefully, you learned from it.