I dont know if I did the problem right check please.

  • Thread starter yaho8888
  • Start date
In summary, The problem asks for the height at which a jet, diving vertically downward at 1200km/hr, must start a quarter turn in order to pull out of the dive without exceeding the maximum acceleration of 4G. Using the equation A=V^2/r, the height is calculated to be approximately 2.834km. To convert the answer to km/hr, all units must be converted to those units before calculation.
  • #1
yaho8888
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0
[SOLVED] I don't know if I did the problem right check please.

Homework Statement


A jet is diving vertically downward at 1200km/hr
the pilot can withstand a maximum acceleration of 4G, before losing consciousness, at what height must the plane start a quarter turn to pull out of the dive? Assume the speed constant.


Homework Equations



A=V^2/r

The Attempt at a Solution



for this circular motion we know that h = radius of the circle
so
4G=(1200km/hr)^2/h
4G=4*9.8m/s=39.2m/s=508032km/hr^2(Not sure if convert it right)
508032=1200^2/h
h=2.834Km
Is this right, the answer has to be in Km so I convert the 4G.
 
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  • #2
If you want the ans in km/hr, convert everything to those units.

4g = v^2/h =>
4*(9.8/1000)/(1/3600)^2 = 1200^2/h. The value of h will be in kms.

EDIT: 1/3600 corrected (1/3600)^2, after being pointed out by yaho8888.
 
Last edited:
  • #3
Looks good; you incorrectly typed m/s instead of m/s^2 in your 4G calculation, but the math came out OK.
 
  • #4
Shooting star said:
If you want the ans in km/hr, convert everything to those units.

4g = v^2/h =>
4*(9.8/1000)/(1/3600) = 1200^2/h. The value of h will be in kms.

wouldn't that be 4*(9.8/1000)/(1/3600)^2
 
  • #5
yaho8888 said:
wouldn't that be 4*(9.8/1000)/(1/3600)^2

Yes. Thanks for pointing out the typo. I'll edit the post.
 

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3. Is it normal to doubt my solution to a problem?

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