Find velocity required to reach a certain height given only height and gravity

In summary, the problem involves finding the initial upward velocity of a mass of gas rising from the sun's surface to a height of 24000km, considering only the acceleration due to gravity. The equations used are a(t) = 0.25, v(t) = 0.25t + C, and s(t) = 0.125t² + Ct + D, with D being assumed to be 0. However, the presence of two variables makes it difficult to solve using substitution or the quadratic formula. The hint is to consider the acceleration to be negative, as it is towards the sun, and to remember the velocity at maximum height.
  • #1
haiku11
11
0
On the surface of the sun the acceleration due to gravity is approx 0.25km/s². A mass of gas forming a solar prominence rises from the sun's surface. If only gravity is considered, what must its initial upward velocity be, if it is to reach a height of 24000km above the surface?

I can't figure out how to do this, I have:
a(t) = 0.25 the integral would give me:
v(t) = 0.25t + C the integral of this would give me:
s(t) = 0.125t² + Ct + D

I'm assuming D would be 0 because that's where the prominence starts on the ground so the equation becomes:
24000 = 0.125t² + Ct
0 = 0.125t² + Ct - 24000

I don't know where to go from here because there are 2 variables and I can't do any substitution using the previous equations. Trying to use the quadratic formula made everything really messy when trying to rearrange it in terms of "t". I even tried doing this the physics way although this is a calculus problem and I still couldn't do it with the 5th motion equation because I would be trying to take the square root of a negative. But if I ignore the negative and take the root I get the approximately right answer of around 109km/s.
 
Physics news on Phys.org
  • #2
haiku11 said:
On the surface of the sun the acceleration due to gravity is approx 0.25km/s². A mass of gas forming a solar prominence rises from the sun's surface. If only gravity is considered, what must its initial upward velocity be, if it is to reach a height of 24000km above the surface?

I can't figure out how to do this, I have:
a(t) = 0.25 the integral would give me:
v(t) = 0.25t + C the integral of this would give me:
s(t) = 0.125t² + Ct + D

I'm assuming D would be 0 because that's where the prominence starts on the ground so the equation becomes:
24000 = 0.125t² + Ct
0 = 0.125t² + Ct - 24000

I don't know where to go from here because there are 2 variables and I can't do any substitution using the previous equations. Trying to use the quadratic formula made everything really messy when trying to rearrange it in terms of "t". I even tried doing this the physics way although this is a calculus problem and I still couldn't do it with the 5th motion equation because I would be trying to take the square root of a negative. But if I ignore the negative and take the root I get the approximately right answer of around 109km/s.

If you are taking s positive upward, your acceleration should be towards the sun, therefore negative. Also, as another hint, what do you know about the velocity when it is at maximum height of 24000?
 
  • #3
Darn it I forgot the acceleration is negative in this case, this changes everything. Thanks.
 

1. What is the formula for finding velocity required to reach a certain height given only height and gravity?

The formula for finding velocity required to reach a certain height given only height and gravity is:v = √(2gh), where v is the velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height.

2. Can the formula be used for any object regardless of its mass?

Yes, the formula can be used for any object regardless of its mass, as long as the acceleration due to gravity remains constant.

3. What units should be used for the height and velocity in the formula?

The height should be measured in meters (m) and the velocity should be measured in meters per second (m/s). It is important to use consistent units in order to get an accurate result.

4. Is there a specific direction for the velocity in this formula?

Yes, the velocity should be in the same direction as the acceleration due to gravity. This means that for objects being thrown upwards, the velocity will be positive, while for objects falling downwards, the velocity will be negative.

5. How can this formula be applied in real-life situations?

This formula can be applied in various real-life situations, such as calculating the velocity of a rocket launching into space, determining the speed of a rollercoaster at the top of a hill, or finding the initial velocity of a projectile being launched at a certain height.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
968
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
Replies
2
Views
713
  • Calculus and Beyond Homework Help
Replies
3
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
8K
  • Calculus and Beyond Homework Help
Replies
10
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top