Question

A quarterback claims that he can throw the football ahorizontal distance of 183 m (200 yd). Furthermore, he claims thathe can do this by launching the

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asked 2021-02-16
A quarterback claims that he can throw the football ahorizontal distance of 183 m (200 yd). Furthermore, he claims thathe can do this by launching the ball at the relatively low angle of 30.0 degree above the horizontal. To evaluate this claim, determinethe speed with which this quarterback must throw the ball. Assumethat the ball is launched and caught at the same vertical level andthat air resistance can be ignored. For comparison, a baseballpitcher who can accurately throw a fastball at 45 m/s (100 mph)would be considered exceptional.

Expert Answers (2)

2021-02-18
For horizontal range of a projectile \(\displaystyle{R}={{V}_{{{0}}}^{{{2}}}}\frac{{\sin{{20}}}}{{g}}\) Hence \(\displaystyle{V}{0}=\sqrt{{{R}\cdot\frac{{g}}{{\sin{{20}}}}}}=\sqrt{{{183}\cdot\frac{{9.8}}{{\sin{{60}}}}}}={45.5}\frac{{m}}{{s}}\)
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Best answer
2021-10-11

Let the velocity of throwing be y,

Consider the vertical motion of football,

We have equation of motion, v = u + at

At maximum height final velocity, v = 0

Initial velocity, \(u = y \sin 34.2 = 0.562 y\)

Acceleration, \(a = -9.81 \ m/s^2\)

Substituting,

\(0=0.562y-9.81t\)

\(t=\frac{0.562y}{9.81}\)

Total time of flight = 2 x Time to reach maximum height

Total time of flight \(=\frac{2 \times 0.562y}{9.81}=0.115y\)

Consider the horizontal motion of football,

We have equation of motion , \(s=ut+0.5at^2\)

Initial velocity , \(u=y \cos 34.2 =0.827y\)

Acceleration, \(a = 0 m/s^2\)

Total time of flight , t=0.115y

Substituting

\(s=0.827y \times 0.115y +0.5 \times 0 \times (0.115y)^2=0.095y^2\)

This is the maximum horizontal distance he can throw,

Given that

\(0.095y^2=161\)

\(y=41.14 \ m/s\)

The speed with which this quarterback must throw the ball = 41.14 m/s

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