At what angle should the bomb be released?

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Homework Help Overview

The discussion revolves around a projectile motion problem involving a plane dropping a bomb on a car. The plane is flying at 200 km/h while the car is traveling at 130 km/h, with the car positioned 78.0 meters below the plane. Participants are exploring the angle at which the bomb should be released.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conversion of units and the drawing of vectors. There is mention of using kinematic equations to determine the time of flight and the distance traveled by the bomb. Some participants question the sufficiency of the problem statement and the relative positions and directions of the plane and car.

Discussion Status

There is a mix of reasoning and calculations presented, with some participants expressing uncertainty about the problem's wording and assumptions. While one participant agrees with a calculated angle of 45 degrees, others raise concerns about the implications of the car being described as "below" the plane and the potential for differing outcomes based on the directions of motion.

Contextual Notes

Participants note that the problem may lack sufficient information and clarity regarding the relative motion of the plane and car, which could affect the interpretation of the solution.

Alpha123
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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.
 
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Alpha123 said:

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?
 
SammyS said:
This looks like a projectile motion problem to me.
SammyS said:
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?
 
Do you know the relative positions of the plane and the car, as well as the relative directions of motion?
 
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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