A football is kicked at ground level with a speed of$18.0\mathrm{m}/\mathrm{s}$at an angle of$31.0°$to the horizontal. How much later does it hit the ground?

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 $\theta ={31}^{\mathrm{o}}$.

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:

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

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.