Chapter 3, Problem 29P

To determine

To find:Whether the football will pass over the bar.

Answer to Problem 29P

The football will not pass over the bar.

Explanation of Solution

Given:

The cross-bar is 3 m above ground and goal post is 36 m away, angle of projection is 37°.

Formula Used:

The kinematics equation is,

  $\begin{array}{l}{S}_y=u_{y}t+\frac{1}{2}a{t}^2\\ R=\frac{u^2\sin 2\theta }{g}\\ v_y=v\sin \theta \\ v_x=u\cos \theta \end{array}$

Calculation:

  $\begin{array}{l}\text{Angle of projection,}\theta =37°\\ \text{Range }R=36\text{ m}\\ \text{Height of cross-bar }H=3\text{ m }\end{array}$

Substituting the values, to find the initial velocity.

  $\begin{array}{l}36=\frac{u^2\sin 2\left(37\right)}{10}\\ u=19.33\text{ m/s}\end{array}$

Substituting the values, to find the final velocity y -component.

  $\begin{array}{l}{v}_y=19.33\sin 37\\ v_y=11.63\text{ m/s}\end{array}$

Substituting the values, to find the final velocity x -component.

  $\begin{array}{l}{v}_x=19.33\cos 37\\ v_x=15.43\text{ m/s}\end{array}$

Displacement along x axis,

  $v_x=\frac{S_x}{t}$

Substituting the values

  $\begin{array}{l}15.43=\frac{36}{6}\\ t=2.33\text{ sec}\end{array}$

Displacement along y axis

  $S_y=u_{y}t-\frac{1}{2}{a}_y{t}^2$

Substituting the values

  $\begin{array}{l}{S}_y=11.63\times 2.33-\frac{1}{2}\times 10\times {\left(2.33\right)}^2\\ S_y=2.27\text{ m}\end{array}$

The distance is less than 3 m. so, the ball will not pass over the bar.

Conclusion:

Thus, the football will not pass over the bar.