A quarterback claims that he can throw the football a horizontal distance of 183

Question

A quarterback claims that he can throw the football a horizontal distance of 183 m (200 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.

________ m/s

Explanation / Answer

same problem with different numbers please have a look at it

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 25° above the horizontal. To evaluate his claim, determine the speed with which this quarterback must throw the ball

Re the baseball reference: A baseball is much smaller and lighter than a football; so the number 45 m/s gives you a ballpark where the absolute fantastic maximal speed of a football might be ...

Let's call the speed of the football V. Then this speed can be factored into a vertical and a horizontal speed, which we call Vv and Vh. By Pythagoras, V = sqrt(Vv^2 + Vh^2).
We also know from trigonometry that tan(25°) = Vv/Vh.
The ball starts to climb with speed Vv; and decelerates with acceleration 9.81m/s². So after Vv/9.81 s, it is at the topmost point; the same time later, it is at the field level again.
During all this time, it flies horizontally Vh * (2 * Vv / 9.81) m, which is supposed to be 196.6 m. So we have

Vh * (2 * Vv / 9.81) = 196.6

From with tan(25°) = Vv/Vh, you get Vv = Vh*tan(25°), so that we can substitute for Vv above:

Vh * (2 * Vh * tan(25°) / 9.81) = 196.6

From this, you can compute Vh; and then Vv; and then (see Pythagoras above) V.

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