Question

An air-plane running on the runway finds the rain falling at an angle of 45° with vertical. The plane is running at a speed of 10 m/s and rain falling vertically. The plane now takes off at angle 15º with horizontal and finds rain falling at same angle with vertical as before relative to it. The speed of the plane after take off is
(A) 10√2 m/s
(B) 10√5 m/s
(C) 10√3 m/s
(D) 10 m/s​

Answer 1

Answer:

The speed of the plane after taking off is found to be $10 \sqrt 2 \;m s^-^1$.

Explanation:

We are given that the vector AB represents the rain falling at a normal, 45° east.

When a man travels at the speed of 10 m/s towards the East.
We now find his velocity relative to the rain.

Now, the vector AD represents the man's travel vector, which is mentioned in the westward direction.

Upon constructing the parallelogram with the vectors, the diagonal works as the resultant vector.

Now, we get the inverted normal (diagonal) to be $10 \sqrt 2 \;m s^-^1$. Thus, the speed of the raindrop can be found to be $14.14 \; ms^-^1$.

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