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 AA,and then it travelled along the curved path ACAC, the hawk flew after the release upto BB for 2s2s,and then followed along straight line ACAC,to catch the mouse 3 m3m 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 AA to CC is 240-3 = 237 m2403=237m

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

237 = 1/2 g t^2237=12gt2 (using, s= 1/2 g t^2s=12gt2)

So, t=6.94 st=6.94s

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

Now,in this time its horizontal displacement (AKAK) will be 19*6.94=131.86 m196.94=131.86m

Now,the question says,after releasing the mouse,the hawk went on flying with its initial velocity for 2s2s,then followed this staright line pathway to catch the mouse at CC,so it must cover pathway ACAC in (6.94-2)=4.94 s(6.942)=4.94s

So,if it dived with velocity vv at an angle thetaθ w.r.t horizontal,then it downward component of velocity is v sin thetavsinθ and horizontal component is v cos thetavcosθ

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

237 = v sin theta * 4.94 + 1/2 g (4.94)^2237=vsinθ4.94+12g(4.94)2 (using, s= ut + 1/2g t^2s=ut+12gt2)

so, v sin theta=23.75vsinθ=23.75

Again in this time the hawk will also have a horizontal displacement of 131.86 m131.86m (as same as that of the mouse)

But it went from AA to BB with its initial velocity,then with velocity v cos thetavcosθ

So, AK = AB + BK = (19*2) +(v cos theta*4.94)=131.86AK=AB+BK=(192)+(vcosθ4.94)=131.86

So, v cos theta =19vcosθ=19

So, v^2 sin^2 theta + v^2 cos ^2 theta = (23.75)^2 + (19)^2v2sin2θ+v2cos2θ=(23.75)2+(19)2

So, v=30.42 ms^-1v=30.42ms1

So, we can say, 30.42*cos theta = 1930.42cosθ=19

or, theta = cos ^-1(19/30.42)=51.35 ^@θ=cos1(1930.42)=51.35