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

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:

$d = Vit + \frac{1}{2}gt^{2}$

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.

$dy = Viyt + \frac{1}{2}gt^{2}$

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

$dy = (0m/s)t + \frac{1}{2}gt^{2}$

                                             $dy = \frac{1}{2}gt^{2}$

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

$dx = Vixt + \frac{1}{2}gt^{2}$

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

$dx = Vixt + \frac{1}{2}(0m/s^{2})t^{2}$

                                                       $dx = Vixt$
--- 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:

$dy = \frac{1}{2}gt^{2}$
$\frac{2dy}{g}=t^{2}$
$\sqrt{\frac{2dy}{g}} = \sqrt{t^{2}}$
                                           $\sqrt{\frac{2dy}{g}} = t$

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

$\sqrt{\frac{2(137m)}{1.63m/s^{2}}} = t$
$\sqrt{\frac{274m}{1.63m/s^{2}}} = t$
$\sqrt{168.098s^{2}}=t$
$12.965s = t$

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. 
$dx = Vixt$
$4,000m = Vix(13s)$
$\frac{4,000m}{13s} = Vix$
$308m/s= Vix$

The horizontal velocity is 308m/s. 

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