How Much Vertical Acceleration Does a Jet Need to Avoid a Hill?

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SUMMARY

The discussion centers on calculating the vertical acceleration required for a jet traveling at 1300 km/h, positioned 35 meters above ground level, to avoid colliding with a 10% slope. The conclusion reached is that a vertical acceleration of 10 m/s² is necessary to prevent impact. The relevant equations mentioned include the position equation x=V and the vertical motion equation y=35+1/2 at². Participants emphasize the importance of understanding the jet's trajectory in relation to the ground slope.

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KukyZ
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New user has been reminded to please always show their work on schoolwork problems.
Homework Statement
A jet travels at speeds of 1300 km/h at 35 m above ground level. Suddenly it meets a slope of 10% and immediately corrects its trajectory accelerating upwards (a(t) constant, in the direction of y).
How much should the acceleration be worth to avoid impact with the ground?
Relevant Equations
x=V
y=35+1/2 at^2
The answer should be 10 m/s^2 but I don't know how to solve it
 
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I guess you meant x=vt.
So you have the position of the jet at time t. What about the position of the ground below it?
 
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KukyZ said:
Homework Statement: A jet travels at speeds of 1300 km/h at 35 m above ground level. Suddenly it meets a slope of 10% and immediately corrects its trajectory accelerating upwards (a(t) constant, in the direction of y).
How much should the acceleration be worth to avoid impact with the ground?
Relevant Equations: x=V
y=35+1/2 at^2

The answer should be 10 m/s^2 but I don't know how to solve it
I suspect that the problem is asking for the smallest possible vertical acceleration that will prevent a collision with the ground. Start by drawing a diagram of the jet's trajectory. The jet must barely graze the hillside.

How do you know that the answer is 10 m/s2?
 
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