What angle did the hawk make with the horizontal during its descent?

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1 Answer
Feb 28, 2018

Please see below

Explanation:

Suppose,the mouse was released at point A,and then it travelled along the curved path AC, the hawk flew after the release upto B for 2s,and then followed along straight line AC,to catch the mouse 3m above the ground.

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Now,when the mouse was released it had an initial horizontal velocity as same as that of the hawk,and due to the pull of the gravity it followed this parabolic path.

So,vertical length from A to C is 2403=237m

So,for falling by this distance if the mouse took time t,then we can write,considering vertical motion only,

237=12gt2 (using, s=12gt2)

So, t=6.94s

this is the time for which the mouse enjoyed free fall,

Now,in this time its horizontal displacement (AK) will be 196.94=131.86m

Now,the question says,after releasing the mouse,the hawk went on flying with its initial velocity for 2s,then followed this staright line pathway to catch the mouse at C,so it must cover pathway AC in (6.942)=4.94s

So,if it dived with velocity v at an angle θ w.r.t horizontal,then it downward component of velocity is vsinθ and horizontal component is vcosθ

So,again considering vertical motion only for the hawk,we can say,

237=vsinθ4.94+12g(4.94)2 (using, s=ut+12gt2)

so, vsinθ=23.75

Again in this time the hawk will also have a horizontal displacement of 131.86m (as same as that of the mouse)

But it went from A to B with its initial velocity,then with velocity vcosθ

So, AK=AB+BK=(192)+(vcosθ4.94)=131.86

So, vcosθ=19

So, v2sin2θ+v2cos2θ=(23.75)2+(19)2

So, v=30.42ms1

So, we can say, 30.42cosθ=19

or, θ=cos1(1930.42)=51.35