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# A quarterback claims that he can throw the football a horizontal distance of 196.6 m (215 yd). Furthermore...?

A quarterback claims that he can throw the football a horizontal distance of 196.6 m (215 yd). Furthermore, he claims that he can do this by launching the ball at the relatively low angle of 35° above the horizontal. To evaluate his claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. For comparison, a baseball pitcher who can accurately throw a fastball at 45 m/s (100 mph) would be considered exceptional. Answer in m/s.

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Let the quarterback throws the ball with velocity u.

As he throws the ball at an angle of 35°, the ball will have two velocity components:

usin35°(vertical) and ucos35°(horizontal). The ball will travel the horizontal distance with the horizontal velocity component only.

Now we have to calculate travel time of the ball before it reaches the ground.

During vertical motion the ball will travel under gravitational force and at the highest point it can climb its velocity will be 0. Putting these in equation we get:

0= usin35°-gt g=Gravitational acceleration= 9.81m/s^2

Or, t=( usin35°/g)

This is only the time for reaching the highest point. The ball will take the same time for reaching the ground from the highest point. Hence total time of travel=2t= (2 usin35°/g)

So total horizontal distance covered by the ball= (2.usin35°)/g × ucos35°=(2u^2 sin35°cos35°)/g=(u^2 sin70°)/g=196.6 m(given)

So, u^2=(196.6×g)/(sin70°)=(196.6×9.81)/(sin70°)

Therefore, u=√((196.6×9.81)/(sin70°) ) = 45.3 m/s

So the velocity with which the person throws the ball is= 45.3 m/s

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