Imagine that a bullet is shot vertically into the air with an initial

In summary: I've been trying for a while and I still can't get it. :(I know it... I've been trying for a while and I still can't get it. :(
  • #1
Art_Vandelay
33
0

Homework Statement



Imagine that a bullet is shot vertically into the air with an initial speed of 9800 m/s. If we ignore air friction, how high will it go?

vi = 9800 m/s
h = ?

Homework Equations



v2 = vi2 + 2aΔy
Ki = Ui --> 1/2mv2 = mgh

F = (Gm1m2)/r


The Attempt at a Solution



Attempt 1

v2 = vi2 + 2aΔy
0 = (9800 m/s)2 + 2(-9.8 m/s2)(Δy)
Δy = 4,900,000 m

Attempt 2

1/2mv2 = mgh
h = 4,900,000 m

Neither of these equations worked, so I tried using gravitational equations, but I couldn't figure it out.
 
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  • #2
Do you know calculus?
What's the actual answer given in the textbook?
 
  • #3
adjacent said:
Do you know calculus?
What's the actual answer given in the textbook?
Unfortunately, I do not know calculus, and there is no answer listed in the back of the book.
 
  • #4
Art_Vandelay said:
Unfortunately, I do not know calculus, and there is no answer listed in the back of the book.

Your answer is correct if we assume that acceleration is constant.Who said it's wrong?

If you don't know calculus, then you should not bother about the varying acceleration..
Edit:It seems you should. What's your grade?
 
Last edited:
  • #5
I typed it into an online program called CAPA, and it replied that my answer was incorrect. I just emailed my professor, and he said that I need to use the full gravitational potential energy equation because the gravitational force is not constant. I've tried a few more attempts, but none of them have been correct.
 
  • #6
Art_Vandelay said:
I typed it into an online program called CAPA, and it replied that my answer was incorrect. I just emailed my professor, and he said that I need to use the full gravitational potential energy equation because the gravitational force is not constant. I've tried a few more attempts, but none of them have been correct.

Please show your attempts then
 
  • #7
Yes, the value you get for Δy should tell you to use the full gravitational force equation since the value you get is of the order of magnitude of the radius of the earth. So the force will be a function of height and you cannot use the approximation U=mgh.
 
  • #8
adjacent said:
Please show your attempts then

Sorry, just got back home.

First, I tried using this equation: 1/2mv2 = (Gm1m2/r2)Δy
Δy = (v2 * r2)/2Gm1
Δy = (98002* 63800002) / (2*(6.67*10-11)* (5.97*1024))
Δy= 4908664 m

Next, I tried omitting the Δy, instead using: r = sqrt(Gm/v2)
r = sqrt((6.67*10-11)* (5.97*1024)/ 98002)
r = 2036 m
 
  • #9
CAF123 said:
Yes, the value you get for Δy should tell you to use the full gravitational force equation since the value you get is of the order of magnitude of the radius of the earth. So the force will be a function of height and you cannot use the approximation U=mgh.

How can I set up my equation so that force is a function of height? I've attempted to work in G and Earth's mass and radius, but I attained an answer too close to my previously found incorrect answers.
 
  • #10
Your first attempt above is incorrect because you are again using the fact that F=mg, which is only true when the height of projectile << radius of earth. Formally, to show that we do a Taylor expansion.

I am not quite sure what you did in the second attempt by simply throwing away some terms.

The method I had in mind does involve calculus as adjacent alluded to. The starting point is to write $$m_1 \ddot{z} = -\frac{Gm_1 m_2}{r^2} = -\frac{Gm_1 m_2}{(r_e + z)^2}$$ and then to do some chain rule and integration.
 
  • #11
CAF123 said:
Your first attempt above is incorrect because you are again using the fact that F=mg, which is only true when the height of projectile << radius of earth. Formally, to show that we do a Taylor expansion.

I am not quite sure what you did in the second attempt by simply throwing away some terms.

The method I had in mind does involve calculus as adjacent alluded to. The starting point is to write $$m_1 \ddot{z} = -\frac{Gm_1 m_2}{r^2} = -\frac{Gm_1 m_2}{(r_e + z)^2}$$ and then to do some chain rule and integration.

Hmm... How would I incorporate the velocity of the object then?

I tried using 1/2mv2 = -(Gm1m2/r), but ended up getting a very small number.
98002 = - (2*(6.67*10-11)*(5.97*1024))/(6.4*106 - h)2

I ended up getting 6402880 m, but I don't want to enter it into the program because I have used almost all of my attempts already.
 
  • #12
The basic idea you want to use is that kinetic energy + potential energy is conserved. Use the correct potential function and it should all fall into place. I'm not a physicist but this is surely how it works.

I think this should not be complicated.
 
  • #13
verty said:
The basic idea you want to use is that kinetic energy + potential energy is conserved. Use the correct potential function and it should all fall into place. I'm not a physicist but this is surely how it works.

I think this should not be complicated.

I know it shouldn't be complicated, and that's making it all the more frustrating. I can't seem to attain the correct answer after spending hours on this one question. I only have two attempts left.

I tried doing 1/2mv2 + mgh = - (GmM/r), so that
1/2(9800)2 + (9.8 m/s2)h = -((6.67*10-11)(5.97*1024)/6380000
h = 1.13 * 107 m (but that is incorrect as well)

I would really appreciate if someone could hold my hand and take me through this because I have no clue what to do.
 
  • #14
Art_Vandelay said:
I tried doing 1/2mv2 + mgh = - (GmM/r), so that
You're mixing up two equations for the the potential energy. Stick to the full one. Hint: r will vary from Rearth to Rearth+h.
 
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  • #15
Try using the full expression for the potential both for the initial and for the turning point. At what r are you at the initial point? At what r are you at the turning point?
 
  • #16
Art_Vandelay said:
I know it shouldn't be complicated, and that's making it all the more frustrating. I can't seem to attain the correct answer after spending hours on this one question. I only have two attempts left.

I tried doing 1/2mv2 + mgh = - (GmM/r)

mgh is the potential energy above the surface if the force of gravity is assumed constant, so you can't use that.

-(gmM/r) is the potential energy of the bullet compared to the potential energy at infinity.
You need the potential energy with respect to the ground.
 
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  • #17
Oooh... So the equation needs to be set up as 1/2mv2 + (-GmM/r) = (-GmM/r+h)

Thank you to everyone who helped! I really appreciate your cooperation with such a dunce as myself! :p
 

1. What factors affect the height and speed of the bullet as it travels through the air?

The main factors that affect the height and speed of a bullet shot vertically into the air include the initial velocity, air resistance, and gravity. The initial velocity, or the speed at which the bullet is shot, will determine how high and far it will travel. Air resistance, which is influenced by the shape and weight of the bullet, will slow down its speed. And gravity will pull the bullet downwards, causing it to eventually fall back to the ground.

2. How does the angle at which the bullet is shot affect its trajectory?

The angle at which the bullet is shot will determine the shape of its trajectory. A bullet shot at a 90-degree angle (perfectly vertical) will travel straight up and then straight back down, while a bullet shot at a smaller angle will have a more curved trajectory. The angle also affects the vertical and horizontal components of the bullet's velocity, which will determine the distance it travels and the height it reaches.

3. What happens to the bullet's speed as it travels higher into the air?

The bullet's speed will decrease as it travels higher into the air. This is due to the force of gravity pulling it down and air resistance slowing it down. As the bullet reaches its maximum height, its speed will momentarily be zero before it begins to fall back to the ground.

4. Can a bullet shot vertically into the air reach an infinite height?

No, a bullet shot vertically into the air cannot reach an infinite height. The bullet's initial velocity, air resistance, and gravity will all play a role in limiting its maximum height. Eventually, the bullet will reach a point where gravity will pull it back down towards the ground.

5. How does the mass of the bullet affect its trajectory?

The mass of the bullet does not significantly affect its trajectory when shot vertically into the air. This is because the force of gravity and air resistance are acting on all objects, regardless of their mass. However, a heavier bullet may experience slightly more air resistance, causing it to slow down faster and have a shorter trajectory compared to a lighter bullet with the same initial velocity.

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