91Ó°ÊÓ

Projectile motion is the path traveled by an object launched in the air and allowed to move under the influence of gravity. For example, kicking a football, firing a cannonball, firing a bullet from the gun, etc.

Suppose the origin is the point from where the football is kicked. Let the positive y-direction be upward.

The angle subtended with the horizontal is $\mathrm{θ}={31}^{\mathrm{o}}$.

Along the vertical direction:

The magnitude of the initial velocity is $v=18.0\mathrm{m}/\mathrm{s}$.

When the football returns to the ground, the displacement is $y=0$.

The acceleration experienced by the football is $a_{\mathrm{y}}=g=9.8\mathrm{m}/{\mathrm{s}}^2$. It is negative because it acts in a downward direction.

For the football, the initial velocity along the vertical direction is given by:

$\begin{array}{c}{v}_{\mathrm{y}}=v\sin \mathrm{θ}\\ =18\sin \mathrm{θ}\\ =\left(18\sin 31°\right)\mathrm{m}/\mathrm{s}\end{array}$

The final velocity of the football along the y-direction is opposite to the initial velocity along the same direction. It is given as:

${v_{\mathrm{y}}}^{\text{'}}=-\left(18\sin 31°\right)\mathrm{m}/\mathrm{s}$

Let the time taken by the football be t. Using the kinematic equation of motion, you get:

$\begin{array}{c}{v_{\mathrm{y}}}^{\text{'}}=v_{\mathrm{y}}+a_{y}t\\ =v_{\mathrm{y}}-gt\\ t=\frac{{v_{\mathrm{y}}}^{\text{'}}-v_{\mathrm{y}}}{-g}\end{array}$

Substitute the known values in the above expression.

$\begin{array}{c}t=\frac{\left(-18\sin 31°\right)\mathrm{m}/\mathrm{s}-\left(18\sin 31°\right)\mathrm{m}/\mathrm{s}}{-9.8\mathrm{m}/{\mathrm{s}}^2}\\ =\frac{-\left(9.27+9.27\right)\mathrm{m}/\mathrm{s}}{-9.8\mathrm{m}/{\mathrm{s}}^2}\\ =1.89\mathrm{s}\end{array}$

Thus, the football hits the ground after 1.89 s.