Question #101710
On the kickoff, the football was in the air for 3 seconds before hitting the ground 68 meters away. How fast did the ball leave the foot of the kicker?
1
Expert's answer
2020-01-27T09:23:08-0500

Time of flight of the football=3=2usinθg(i)3=\dfrac{2u\sin\theta}{g}-----(i)


68=ucosθ×T68=u\cos\theta \times T


68=ucosθ×368=u \cos\theta \times 3


u=683cosθ(ii)u=\dfrac{68}{3\cos\theta}--------(ii)


From the equation (i) and (ii)


3=2×68sinθ3cosθ3=\dfrac{2\times 68 \sin\theta}{3\cos\theta}


tanθ=9138\tan \theta=\dfrac{9}{138}


θ=3.731\theta=3.731



68=ucos(3.7)×368=u\cos (3.7) \times 3

u=683cos3.7=680.99×3=22.89m/secu=\dfrac{68}{3\cos{3.7}} =\dfrac{68}{0.99\times 3}=22.89m/sec


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