Chapter 9, Problem 91RE

To determine

Calculate the speed of the football when it is released.

Answer to Problem 91RE

The speed of the football is $54.1ft/\sec$ .

Explanation of Solution

Given: It is given in the question that the quarterback of a football team releases a pass at a height $7$ feet above the playing field, and the football is caught at a height of $4$ feet, $30$ yards directly downfield. The pass is released at an angle of ${35}^{\circ }$ with the horizontal. The parametric equations for the path of the football are given by $x=0.82v_{{}^{\circ }}\mathrm{t}andy=7+0.57v_{{}^{\circ }}t-t^2$ , where $v_{{}^{\circ }}$ is the speed of the football (in feet per second) when it is released.

Concept Used:

In this, use the concept of projectile motion and understand the question and do the answer.

Calculation:

Here, the two equations are $x=0.82v_{{}^{\circ }}\mathrm{t}andy=7+0.57v_{{}^{\circ }}t-t^2$

y(t) = number of feet above ground at t seconds.

  $30yards=90feet$ .At this time, $4=y(\frac{90}{v_{{}^{\circ }}\cos {35}^{{}^{\circ }}})$

  $\begin{array}{l}{v}_{{}^{\circ }}\cos {35}^{{}^{\circ }}t=90\\ t=\frac{90}{v_{{}^{\circ }}\cos {35}^{{}^{\circ }}}\end{array}$

Put that value in the second equation,

  $\begin{array}{l}$ 4=7+v_{{}^{\circ }}\sin {35}^{{}^{\circ }}(\frac{90}{v_{{}^{\circ }}\cos {35}^{{}^{\circ }}})-16{(\frac{90}{v_{{}^{\circ }}\cos {35}^{{}^{\circ }}})}^2$$ -3=v_{{}^{\circ }}\times 0.57(\frac{90}{v_{{}^{\circ }}\times 0.82})-16{(\frac{90}{v_{{}^{\circ }}\times 0.82})}^2$v_{\circ }=54.1ft/\sec \end{array}$

Conclusion:

The speed of the football is $54.1ft/\sec$ .