Question

2021-02-18

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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