Question

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?

Short answer

The football hits the ground after 1.89 s.

Step-by-step solution

Step 1. Meaning of projectile motion

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.

Step 2. Assumptions for the motion of the ball

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

Step 3. Identification of the given data

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.

Step 4. Determination of the initial velocity of the football

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

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

Step 5. Determination of the final velocity of the football

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

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

Step 6. Determination of the time taken by the football to reach the ground

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.