A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball?
Solution:
To illustrate the problem, consider the following figure:
Part A
Since the starting position has the same elevation as when it hits the ground, the speeds at these points are the same. The final speed is computed by solving the resultant of the horizontal and vertical velocities. That is
\begin{align*}
\text{v}_{\text{f}} & =\sqrt{\left(\text{v}_{\text{ox}}\right)^2+\left(\text{v}_{\text{oy}}\right)^2} \\
\text{v}_{\text{f}} & =\sqrt{\left(16\:\text{m/s}\right)^2+\left(12\:\text{m/s}\right)^2} \\
\text{v}_{\text{f}} & =\sqrt{400\:\text{(m/s)}^2} \\
\text{v}_{\text{f}} & =20\:\text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Part B
Consider the two points: (1) the starting point and (2) the highest point.
We know that at the highest point, the vertical velocity is zero. We also know that the total time of the flight is twice the time from the beginning to the top.
So, we shall use the formula \text{t}=\frac{\text{v}_{\text{f}}-\text{v}_{\text{o}}}{\text{a}}.
\begin{align*}
\text{t} & =2\left(\frac{\text{v}_{\text{top}}-\text{v}_{\text{o}}}{\text{a}}\right) \\
\text{t} & =2\left(\frac{0\:\text{m/s}-12\:\text{m/s}}{-9.81\:\text{m/s}^2}\right) \\
\text{t} & =2.45\:\text{s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Part C
The maximum height attained can be calculated using the formula \left(\text{v}_{\text{f}}\right)^2=\left(\text{v}_{\text{o}}\right)^2+2\text{a}\text{y}.
The maximum height is calculated as follows:
\begin{align*}
\left(\text{v}_{\text{f}}\right)^2 & =\left(\text{v}_{\text{o}}\right)^2+2\text{ay} \\
\text{y}_{\max } & =\frac{\left(\text{v}_{\text{top}}\right)^2-\left(\text{v}_{\text{o}}\right)^2}{2\text{a}} \\
\text{y}_{\max }& =\frac{\left(0\:\text{m/s}\right)^2-\left(12\:\text{m/s}\right)^2}{2\left(-9.81\:\text{m/s}^2\right)} \\
\text{y}_{\max } & =7.34\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}