What Is the Maximum Initial Velocity of a WW2 German U2 Rocket?

In summary: Don't you just multiply the time it took to reach the max height by 2 to get the time for the entire motion. That's what I did, and then I used that time value in my calculations for the x component.In summary, the conversation discusses calculating the maximum initial velocity of a German V-2 rocket from WW2, which had a range of 300km and reached a maximum height of 100km. The conversation includes the use of x and y component variables and equations, such as the big 5 kinematics equations, to determine the initial velocity. However, the output of pythagorean theorem and other equations did not match the expected answer, likely due to using incorrect units in the calculations. The conversation also
  • #1
Balsam
226
8

Homework Statement


A German U2 rocket from WW2 had a range of 300km, reaching a maximum height of 100km. Determine the rocket's maximum intial velocity.

X component variables: di=0, df=300, a=0, time=9, vi=?
Y component variables: a=-9.8, vf=0, vi=?, di=0, df=100, displacement=100, Note: For the y components, I counted the maxium height as the final position and the velocity at the max height as the final velocity.

Homework Equations


The big 5 kinematics equations

The Attempt at a Solution


I attempted to find the initial velocity for the x and y components and then use pythagorean theorem to solve for the magnitude of the max velocity as you would solve for the hypotenuse of a triangle and then I would use the cosine law to solve for the angle at which the rocket was launched. But, I got the incorrect magnitude after doing pythagorean theorem, so I stopped. Here's what I did:

First, I calculated initial velocity for the y component, using the equation vf^2=vi^2+2a(displacement) --- I got vi=~44.3m/s. Then, I used another equation with the y component variables to solve for time because I needed another variable to use for the x component equations. For finding the time, I used displacement=1/2(vf+vi)(time). I got time=4.5s, but since the variables I used were variables only from the initial position to the max height, I think that using those variables only solved for the time up to the max height, so I multiplied by 2 to get the time for the whole motion--- 9 s. Then, I used the equation df=di+vi(time)+1/2a(t)^2. I got vi=33.3.Then, I used pythagorean theorem with the magnitudes of my x and y component velocities---- a^2=b^2-c^2 -- I plugged in the x and y component velocities for b and c and solved for a. I got a=55.4. This should've been equal to the magnitude of the max velocity, but the book says the answer is 1.75 x 10^3 m/s. Can someone tell me where I went wrong.
 
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  • #2
Balsam said:

Homework Statement


A German U2 rocket from WW2 had a range of 300km, reaching a maximum height of 100km. Determine the rocket's maximum intial velocity.

X component variables: di=0, df=300, a=0, time=9, vi=?
Y component variables: a=-9.8, vf=0, vi=?, di=0, df=100, displacement=100, Note: For the y components, I counted the maxium height as the final position and the velocity at the max height as the final velocity.

Homework Equations


The big 5 kinematics equations

The Attempt at a Solution


I attempted to find the initial velocity for the x and y components and then use pythagorean theorem to solve for the magnitude of the max velocity as you would solve for the hypotenuse of a triangle and then I would use the cosine law to solve for the angle at which the rocket was launched. But, I got the incorrect magnitude after doing pythagorean theorem, so I stopped. Here's what I did:

First, I calculated initial velocity for the y component, using the equation vf^2=vi^2+2a(displacement) --- I got vi=~44.3m/s. Then, I used another equation with the y component variables to solve for time because I needed another variable to use for the x component equations. For finding the time, I used displacement=1/2(vf+vi)(time). I got time=4.5s, but since the variables I used were variables only from the initial position to the max height, I think that using those variables only solved for the time up to the max height, so I multiplied by 2 to get the time for the whole motion--- 9 s.Then, I used the equation df=di+vi(time)+1/2a(t)^2. I got vi=33.3.Then, I used pythagorean theorem with the magnitudes of my x and y component velocities---- a^2=b^2-c^2 -- I plugged in the x and y component velocities for b and c and solved for a. I got a=55.4. This should've been equal to the magnitude of the max velocity, but the book says the answer is 1.75 x 10^3 m/s. Can someone tell me where I went wrong.

For one thing, in your initial calculations, you apparently used a range of 300 meters, instead of 300 kilometers.
Ditto for the maximum altitude; you used 100 meters instead of 100 kilometers.

It makes a difference, those pesky units. :rolleyes:BTW, the name of the rocket was the V-2, not the U-2. The U-2 was a famous U.S. jet spy plane.
 
  • #3
SteamKing said:
For one thing, in your initial calculations, you apparently used a range of 300 meters, instead of 300 kilometers.
Ditto for the maximum altitude; you used 100 meters instead of 100 kilometers.

It makes a difference, those pesky units. :rolleyes:BTW, the name of the rocket was the V-2, not the U-2. The U-2 was a famous U.S. jet spy plane.
My textbook says U2, but okay. And thank you.
 
  • #4
  • #5
SteamKing said:
Another quality publication. I wonder what other whoppers lurk between its covers.

https://en.wikipedia.org/wiki/V-2_rocket

Was I right about multiplying the time by 2 for the y component?
 
  • #6
Balsam said:
Was I right about multiplying the time by 2 for the y component?
You got bigger fish to fry than that. Did you not read the part about using the wrong units in your calculations?
 
  • #7
SteamKing said:
You got bigger fish to fry than that. Did you not read the part about using the wrong units in your calculations?

Yes, but I can easily fix that, the time would also affect my calculations
 
  • #8
Balsam said:
Yes, but I can easily fix that, the time would also affect my calculations
Once the rocket reaches maximum altitude, it's no longer powered. You can calculate the time it takes to free-fall back to earth.
 
  • #9
SteamKing said:
Once the rocket reaches maximum altitude, it's no longer powered. You can calculate the time it takes to free-fall back to earth.

Don't you just multiply the time it took to reach the max height by 2 to get the time for the entire motion. That's what I did, and then I used that time value in my calculations for the x component.
 
  • #10
Balsam said:
Don't you just multiply the time it took to reach the max height by 2 to get the time for the entire motion. That's what I did, and then I used that time value in my calculations for the x component.
At launch, the rocket has zero initial velocity, and it takes time for the rocket to build up to its maximum velocity before the engine cuts off. That's why I suggested calculating the time to free fall from the highest altitude, since it's not clear the velocity profile going up while powered will be the same one coming down in free-fall.
 
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