91Ó°ÊÓ

A projectile that is projected from a height yabove the ground with initial velocity${\stackrel{→}{v}}_0$follows a parabolic path. If the point of projection is taken as the origin and the upward direction is taken as positive y-axis, then the kinematics equations for projectile motion can be written as:

$\begin{array}{c}{v}_{\mathrm{x}}=v_{\mathrm{x}0}\\ x=v_{\mathrm{x}0}t\end{array}$

$\begin{array}{c}{v}_{\mathrm{y}}=v_{\mathrm{y}0}-gt\\ y=v_{\mathrm{y}0}t-\frac{1}{2}g{t}^2\\ v_{\mathrm{y}}^2=v_{\mathrm{y}0}^2-2gy\end{array}$

Here, xandy are the horizontal and vertical displacements of the projectile traveled in time, $v_{\mathrm{x}0}$ and $v_{\mathrm{y}0}$ are the horizontal and vertical components of the initial velocity of the particle, and $v_{\mathrm{x}},v_{\mathrm{y}}$are the horizontal and vertical components of the velocity of the particle at time t.

The football leaves the ground at an angle,$\mathrm{θ}=37.0°$

The initial velocity of the football,$v_0=20.0\mathrm{m}/\mathrm{s}$

The height of the crossbar,$h=3.05\mathrm{m}$

The horizontal distance between the initial point and the goalposts is$x=36.0\mathrm{m}$

Consider the point of kicking the football as the origin and the upward direction as the positive y-direction. Therefore, the acceleration of the football will be equal to the negative of the acceleration due to gravity, i.e.,$a=-g$.

Let $v_{\mathrm{x}0}$and $v_{\mathrm{y}0}$be the horizontal and vertical components of the initial velocity ${\stackrel{→}{v}}_0$of the projectile. You can write the values of the horizontal and vertical components of the initial velocity ${\stackrel{→}{v}}_0$of the projectile in terms of $\mathrm{θ}$in the following manner:

$\begin{array}{c}{v}_{\mathrm{x}0}=v_0\cos 37.0°\\ =\left(20.0\mathrm{m}/\mathrm{s}\right)\left(0.799\right)\\ =16.0\mathrm{m}/\mathrm{s}\end{array}$

$\begin{array}{c}{v}_{\mathrm{y}0}=v_0\sin 37°\\ =\left(20.0\mathrm{m}/\mathrm{s}\right)\left(0.602\right)\\ =12.0\mathrm{m}/\mathrm{s}\end{array}$

Let t be the time taken to cover the horizontal distance of 36.0 m.

Using the kinematic equation for the horizontal motion of the football, you will get time tas:

$\begin{array}{c}x=v_{x0}t\\ t=\frac{x}{v_{x0}}\\ =\frac{\left(36.0\mathrm{m}\right)}{\left(16.0\mathrm{m}/\mathrm{s}\right)}\\ =2.25\mathrm{s}\end{array}$

This is the time required by the projectile to reach the goalposts.

Let y be the vertical distance traveled by the football at time t.

Using the kinematic equation for the vertical motion of the projectile, you will get:

$\begin{array}{c}y=v_{\mathrm{y}0}t-\frac{1}{2}g{t}^2\\ y=\left(12.0\mathrm{m}/\mathrm{s}\right)\left(2.25\mathrm{s}\right)-\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(2.25\mathrm{s}\right)}^2\\ =2.19\mathrm{m}\end{array}$

Thus, the vertical distance covered by the football at the goalposts is 2.19 m. However, the height of the goalposts is 3.05 m. Therefore, the football will not pass over the bar and thus will not make a field goal.

The football will cross the height equal to the height of the goalposts twice as it follows a parabolic path.Let t be the time at which it crosses the height of the goalposts.

Using the kinematic equation for the vertical motion of the football, you will get:

$\begin{array}{c}h=v_{\mathrm{y}0}t\text{'}-\frac{1}{2}g{\left(t\text{'}\right)}^2\\ 3.05=\left(12.0\right)t\text{'}-\frac{1}{2}\left(9.8{\mathrm{ms}}^{-2}\right){\left(t\text{'}\right)}^2\\ 4.9{\left(t\text{'}\right)}^2-\left(12.0\right)t\text{'}+3.05=0\end{array}$

This is a quadratic equation whose solutions are:

$\begin{array}{c}t\text{'}=\frac{\left(12.0\right)Â\pm \sqrt{{\left(12.0\right)}^2-4\left(4.9\right)\left(3.05\right)}}{2×4.9}\\ =\frac{\left(12.0\right)Â\pm \left(9.18\right)}{9.8}\end{array}$

The above expression gives two values of time, which are as follows:

$\begin{array}{c}{t\text{'}}_1=\frac{\left(12.0\right)+\left(9.18\right)}{9.8}\\ =2.16\mathrm{s}\\ {t\text{'}}_2=\frac{\left(12.0\right)-\left(9.18\right)}{9.8}\\ =0.29\mathrm{s}\end{array}$

Thus, the football will cross the height which is equal to the height of the goalposts twice.

Using the kinematic equation for the horizontal motion of the football, you will get:

$\begin{array}{c}{x}_1=v_{\mathrm{x}0}{t\text{'}}_1\\ =\left(16.0\mathrm{m}/\mathrm{s}\right)\left(2.16\mathrm{s}\right)\\ =34.6\mathrm{m}\\ x_2=v_{\mathrm{x}0}{t\text{'}}_2\\ =\left(16.0\mathrm{m}/\mathrm{s}\right)\left(0.29\mathrm{s}\right)\\ =4.6\mathrm{m}\end{array}$

Thus, the football will cross the height equal to the height of the goalposts twice at distances 34.6 m and 4.6 m. Therefore, you can say that the football will make a field goal and score if it is kicked between 4.6 m to 34.6 m from the goalposts.