Physics, 17.10.2019 19:30 Babycute8279

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

Answers: 3

Show answers

Answers

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.

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

--- 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

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

$\frac{4,000m}{13s} = Vix$
308m/s= Vix

The horizontal velocity is 308m/s.

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

Answer from: Quest

b.transmitters send radio waves, and receivers capture radio waves.

Answer from: Quest

If football was tossed in air without gravity or friction the ball would go into space

Another question on Physics

Physics, 22.06.2019 06:50

What is the stall speed at sea level of an airplane that weights 10000 lbs., has a wing area of 300ft^2, and a maximum lift coefficient of 1.4? what is the stall speed if flaps that double cl,max are applied?

Answers: 1

Answer

Physics, 22.06.2019 17:20

Could we see a galaxy that is 20 billion light-years away?

Answers: 1

Answer

Physics, 22.06.2019 17:40

Aball is thrown horizontally from a cliff at 8m/s .in what direction is the ball moving 2s later?

Answers: 1

Answer

Physics, 22.06.2019 19:30

Visualize the problem and identify special cases first examine the problem by drawing a picture and visualizing the motion. apply newton's 2nd law, ∑f⃗ =ma⃗ , to each body in your mind. don't worry about which quantities are given. think about the forces on each body: how are these consistent with the direction of the acceleration for that body? can you think of any special cases that you can solve quickly now and use to test your understanding later? one special case in this problem is if m2=0, in which case block 1 would simply fall freely under the acceleration of gravity: a⃗ 1=−gj^.

Answers: 1

Answer

You know the right answer?

While attempting a landing on the moon, astronauts had to change their landing site and land at a sp...

Questions