Does the plane strike the ground?

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The discussion revolves around a physics problem involving a plane flying horizontally at 1300 km/h and an upward slope of 7.5 degrees. The initial height of the plane is 76 meters, and the main challenge is determining when the plane will strike the ground. Participants clarify that gravity is not necessary for solving this problem, emphasizing the use of trigonometry to form a right triangle with the given dimensions. One participant initially miscalculates the plane's speed but later corrects it to 361.11 m/s. The final solution indicates that the plane strikes the ground in approximately 1.60 seconds.
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Homework Statement


A pilot flies horizontally at 1300 km/h, at height h = 76 m above initially level ground. However, at time t = 0, the pilot begins to fly over ground sloping upward at angle θ = 7.5° (see the figure). If the pilot does not change the airplane's heading, at what time t does the plane strike the ground?
upload_2015-8-29_22-13-0.png

Homework Equations


upload_2015-8-29_22-13-11.png


The Attempt at a Solution


Second question today, been taking chunks at my homework all throughout the day!

So i know the Xi is 76 m. I know that the Vi is 36.11 m/s once i convert from kilometers. I know my t=0 and that since the plane is falling, my a = 9.8 m/s^2. I am not sure what the angle 7.5 does to the problem however. That is where I am stuck.
 
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Why do you say the plane is falling? The question says it is flying horizontally.

There is no need to bring gravity into this question. It's just trigonometry. You have a right triangle with the two non-hypotenuse sides 76m and x metres where x is the distance flown between t=0 and the crash, and the angle is 7.5 degrees. So you can work out x and then work out t from that using the plane's speed.

By the way, your calc of speed in m/s is wrong. 10m/s is 36 km/h. So 1300 km/h is going to be an awful lot more than 36.11 m/s
 
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Thank you for your response!
It seems i screwed up a basic conversion, what I meant to say that it was 361.11 m/s.'

EDIT:
However, i am still unsure where to take this problem with the information listed above.
 
Last edited:
Was there something about the following you didn't understand?

andrewkirk said:
There is no need to bring gravity into this question. It's just trigonometry. You have a right triangle with the two non-hypotenuse sides 76m and x metres where x is the distance flown between t=0 and the crash, and the angle is 7.5 degrees. So you can work out x and then work out t from that using the plane's speed.

Can you draw a diagram of this triangle?
 
I worked through the problem again and understood what you were saying earlier :) I solved for the adjacent and then did delta x= vi*t+1/2(a*t^2). ended up with 1.60 s as the final answer and was right. Thank you :3.
 
Drub said:
I solved for the adjacent and then did delta x= vi*t+1/2(a*t^2).

That equation is somewhat overkill for this problem. Easier to use..

velocity = distance/time
or
time = distance/velocity.
 
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