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.
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 m240−3=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 m19⋅6.94=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(6.94−2)=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=vsinθ⋅4.94+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=(19⋅2)+(vcosθ⋅4.94)=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=30.42ms−1
So, we can say, 30.42*cos theta = 1930.42⋅cosθ=19
or, theta = cos ^-1(19/30.42)=51.35 ^@θ=cos−1(1930.42)=51.35∘