Answer to Question #101710 in Physics for Rosa

Question

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?

Answer 1

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

$68=u\cos\theta \times T$

$68=u \cos\theta \times 3$

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

From the equation (i) and (ii)

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

$\tan \theta=\dfrac{9}{138}$

$\theta=3.731$

$68=u\cos (3.7) \times 3$

$u=\dfrac{68}{3\cos{3.7}} =\dfrac{68}{0.99\times 3}=22.89m/sec$