Chapter 9, Problem 94RE

To determine

Find the time at which the receiver has to position himself after the quarterback releases the ball.

Answer to Problem 94RE

The time for the receiving is $2\sec$ .

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. Use the equation of x − axis to get the time.

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, $4=y(\frac{90}{v_{{}^{\circ }}\cos {35}^{{}^{\circ }}})$

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

  $\begin{array}{l}t=\frac{90}{54.1\times 0.87}\\ t=\frac{90}{47.06}\\ t=1.9\approx 2\sec .\end{array}$

Conclusion:

The time is $2\sec$ .