Time of flight of the football=3=2usinθg−−−−−(i)3=\dfrac{2u\sin\theta}{g}-----(i)3=g2usinθ−−−−−(i)
68=ucosθ×T68=u\cos\theta \times T68=ucosθ×T
68=ucosθ×368=u \cos\theta \times 368=ucosθ×3
u=683cosθ−−−−−−−−(ii)u=\dfrac{68}{3\cos\theta}--------(ii)u=3cosθ68−−−−−−−−(ii)
From the equation (i) and (ii)
3=2×68sinθ3cosθ3=\dfrac{2\times 68 \sin\theta}{3\cos\theta}3=3cosθ2×68sinθ
tanθ=9138\tan \theta=\dfrac{9}{138}tanθ=1389
θ=3.731\theta=3.731θ=3.731
68=ucos(3.7)×368=u\cos (3.7) \times 368=ucos(3.7)×3
u=683cos3.7=680.99×3=22.89m/secu=\dfrac{68}{3\cos{3.7}} =\dfrac{68}{0.99\times 3}=22.89m/secu=3cos3.768=0.99×368=22.89m/sec
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