An airplane is taking off at 132 m/s, 28.2 degrees above the horizontal.?

Question

1- What length (in m) of the horizontal runway does it travel over in 7.62 seconds?
2- How many seconds are required for the airplane to gain 1.00 km of altitude?

Answer 1

Horizontal velocity = 132cos28.2
Vertical velocity = 132cos28.2

1) Distance = speed X time
=132cos28.2 X 7.62 = 886.4502642 m.

2) 1Km =1000 m.
Distance = speed X time
1000=132sin28.2 X Time
TIme =16.03162695 s.

Answer 2

i am assuming this is a question for a high school or college physics class, so let me say to your teacher.......as a pilot with over 1600 hours, let me say that any angle of attack over 15 degrees in any private or commercial aircraft will likely result in airflow sepration and a stall and crash...so whoever wrote the question should at least know their basic aerodynamics before making up unrealistic questions....unless we are talking about military aircraft which have a much higher thust/weight ratio...not the answer you were looking for but you may want to add it into your answer to the question because the problem is unrealistic....also angle of attack decreases with altitude, so you would not be able to hold an angle of attack of almost 30degrees to that altitude anyway

Answer 3

Why are you asking a bunch of random Yahoo folk to do your Physics homework for you?

(hint: use the cosine function to figure out #1, use the sine function for #2)

Answer 4

1. In the horizontal
x(t)=132*cos(28.2)*t
x(7.62)=132*cos(28.2)*7.62
886 meters


2. In the vertical
y(t)=132*sin(28.2)*t
y(t)=1000=132*sin(28.2)*t
solve for t
16 seconds
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