At what angle should the bomb be released?

  • #1
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Homework Statement


There is a plane flying at 200 km/h and wants to drop a bomb on a car traveling 130 km/h which is 78.0m below the plane. at what angle should the bomb be released? the answer is 45 degrees.

Homework Equations


sine law: SinA/a= SinB/b
cosine law: a^2=b^2+c^2+bcCosA
v=d/t
sine,cosine and tan trig ratios

The Attempt at a Solution


converted 78.0m to 0.078 km.
I have drawn out the vectors but have not been able to add them. Any help would be appreciated.
 

Answers and Replies

  • #2
SammyS
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Homework Statement


There is a plane flying at 200 km/h and wants to drop a bomb on a car traveling 130 km/h which is 78.0m below the plane. at what angle should the bomb be released? the answer is 45 degrees.

Homework Equations


sine law: SinA/a= SinB/b
cosine law: a^2=b^2+c^2+bcCosA
v=d/t
sine,cosine and tan trig ratios

The Attempt at a Solution


converted 78.0m to 0.078 km.
I have drawn out the vectors but have not been able to add them. Any help would be appreciated.
It seems to me that there is not sufficient information to answer this.

Have you stated the entire problem?

Is there a figure to go with this?
 
  • #3
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This looks like a projectile motion problem to me.
This looks like a projectile motion problem to me.
i subtracted the velocities to get 70km/h which converts to 19.44m/s
then i plugged it into equation equation d=v1t + 1/2 at^2
78.0m = 1/2(9.8 m/s/)t^2
t= 3.99s

then with the t i found the distance between the plane and car that the bomb travels.
d=vt
=(19.44m/s)(3.99s)
=77.57m

then i found the angle tanx= 78/77.57
x=45 degrees
is this correct?
 
  • #4
SammyS
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Do you know the relative positions of the plane and the car, as well as the relative directions of motion?
 
  • #5
CWatters
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i subtracted the velocities to get 70km/h which converts to 19.44m/s
then i plugged it into equation equation d=v1t + 1/2 at^2
78.0m = 1/2(9.8 m/s/)t^2
t= 3.99s

then with the t i found the distance between the plane and car that the bomb travels.
d=vt
=(19.44m/s)(3.99s)
=77.57m

then i found the angle tanx= 78/77.57
x=45 degrees
is this correct?
I agree with your answer.

However I do think the problem statement is badly worded - for example it says the car is "below" the aircraft which isn't the case. The answer also assumes the aircraft and car are going in the same direction. If they are going in opposite directions you get a different answer.
 

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