# Thanks for the help with my previous problems.

The acceleration to gravity on the moon is 1.6m/sec^2.

If a rock is dropped into a crevasse, how fast it will be going just before it hits bottom 30 sec later?

How far below the point of release is the bottom of the crevasse?

If instead of being released from the rest, the rock is thrown into the crevasse from the same point with a downward velocity of 4 m/s, when will it hit the botoom and how fast will it be going when it does.

Any guidance is appreciated.

Formulas:

vfinal= vinitial + acceleration*time

finalposition= initialposition + vinitial*time + 1/2 acceleration*time^2

The acceleration to gravity on the moon is 1.6m/sec^2.** acceleration is given, downward, lets call that the negative direction. acceleration = -1.6m/s^2**

If a rock is dropped into a crevasse, how fast it will be going just before it hits bottom 30 sec later?** initial velocity= 0. Time is 30 sec.
veloictyfinal= initialvelocity + acceleration *time.
If you did this right, the velocity final is negative, meaning it is downward**

How far below the point of release is the bottom of the crevasse?

**finalposition= initialposition + vinitial*time + 1/2 acceleration*time^2**

If instead of being released from the rest, the rock is thrown into the crevasse from the same point with a downward velocity of 4 m/s, when will it hit the botoom and how fast will it be going when it does.

**here initial velocity is -4m/s.**

Use the distance found in the second part, the distance in the crevice cant change. Solve for time (note: it is a quadratic, use quadratic equation). Once time is found, solve for the final velocity.

Use the distance found in the second part, the distance in the crevice cant change. Solve for time (note: it is a quadratic, use quadratic equation). Once time is found, solve for the final velocity.

For the last part, do we use the same formula, that is vfinal= vinitial + acceleration*time, to get the quadratic equation?

No. You know vinitial, you know acceleration, and you just found time. So it is a plug and chug.

So I find time using the formula, v = d/t ?

That is t = d/v = 720/4 = 180

And v(final) = v(initial) + acceleration * time^2

v(final) = -4 + 1.6 * (180)^2

I am lost!

I am supposed to get 27.60 sec & 48.166 m/s

Find time using the quadratic you just had. Once time is found,

vfinal=vinitial+ acceleraiton*time.

Memorize those two formulas I started you out with. Use flashcards. In this last post, you typed it wrong (see above), you cant afford that. Memorize them now.

But there are two unknowns in the equation, right? time and v(final). How do you get the quadratic?

12 years ago

9 months ago

## Apologies for the confusion. Let's break it down step by step.

Given:

Acceleration (due to gravity on the moon) = -1.6 m/s^2

Initial velocity (vinitial) = -4 m/s

Distance (d) = 720 m

Step 1: Calculate time (t) using the formula d = vinitial * t + (1/2) * acceleration * t^2.

720 = -4 * t + (1/2) * (-1.6) * t^2

Step 2: Rearrange the equation to form a quadratic equation.

(1/2) * (-1.6) * t^2 - 4 * t + 720 = 0

Step 3: Solve the quadratic equation to find the values of t.

Using the quadratic formula, t = (-b Â± âˆš(b^2 - 4ac)) / (2a), where a = (1/2) * (-1.6), b = -4, and c = 720.

Calculating the values, t = (-(-4) Â± âˆš((-4)^2 - 4 * (1/2) * (-1.6) * 720)) / (2 * (1/2) * (-1.6))

t = (4 Â± âˆš(16 + 1152)) / (-0.8)

t = (4 Â± âˆš1168) / (-0.8)

Step 4: Calculate the two possible values of t.

t â‰ˆ (4 Â± 34.192) / (-0.8)

t â‰ˆ -42.74 / -0.8 â‰ˆ 53.425 sec (negative value is not considered)

t â‰ˆ 38.74 / -0.8 â‰ˆ -48.166 sec (negative value is not considered)

Step 5: Calculate the final velocity (vfinal) using vfinal = vinitial + acceleration * time.

For t â‰ˆ 53.425 sec:

vfinal = -4 + (-1.6) * 53.425

vfinal â‰ˆ -4 - 85.480

vfinal â‰ˆ -89.480 m/s

For t â‰ˆ -48.166 sec (disregarded since negative time is not valid):

vfinal = -4 + (-1.6) * (-48.166)

vfinal â‰ˆ -4 + 77.066

vfinal â‰ˆ 73.066 m/s

Therefore, the correct values are:

Time to hit the bottom of the crevasse â‰ˆ 53.425 sec

Final velocity just before hitting the bottom â‰ˆ -89.480 m/s

9 months ago

## Apologies for the confusion. You are correct, there are two unknowns in the equation, time and v(final). To solve for both of them, we need to use the following formula:

finalposition = initialposition + vinitial * time + 1/2 * acceleration * time^2

Since we know the initial position (which is 0 since it's dropped from rest), the initial velocity (which is -4 m/s), and the acceleration (which is -1.6 m/s^2), we can substitute these values into the formula:

finalposition = 0 + (-4) * time + 1/2 * (-1.6) * time^2

Simplifying this equation will give you a quadratic equation:

finalposition = -4t - 0.8t^2

Now, we also know that the final position is the bottom of the crevasse, which means it is 30 seconds below the point of release. So we can set the final position to -30 and solve for time:

-30 = -4t - 0.8t^2

Rearranging this equation will give you the quadratic equation form:

0.8t^2 + 4t - 30 = 0

To solve this quadratic equation, you can use the quadratic formula:

t = (-b Â± âˆš(b^2 - 4ac)) / 2a

where a = 0.8, b = 4, and c = -30. Plugging in these values into the formula will give you the two possible solutions for time.

Once you have found the value of time, you can substitute it back into the equation:

v(final) = v(initial) + acceleration * time

v(final) = -4 + (-1.6) * time

This will give you the final velocity of the rock when it hits the bottom of the crevasse.