Calculating Projectile Motion: Hot Rock Projectile

Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for $t$ first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain $v$ and ${\theta }_v$ at the final time $t$ determined in the first part of the example.

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

If we take the initial position $y_0$ to be zero, then the final position is $y=-\text{20}\text{.0 m}\text{.}$ Now the initial vertical velocity is the vertical component of the initial velocity, found from $v_{0y}=v_0\text{sin}{\theta }_0$ = ($\text{25}\text{.}\text{0 m/s}$)($\text{sin 35.0º}$) = $\text{14}\text{.}\text{3 m/s}$. Substituting known values yields

Rearranging terms gives a quadratic equation in $t$:

This expression is a quadratic equation of the form ${\mathrm{at}}^2+\mathrm{bt}+c=0$, where the constants are $a=4.90$, $b=–14.3$, and $c=–\mathrm{20.0.}$ Its solutions are given by the quadratic formula:

This equation yields two solutions: $t=3.96$ and $t=–1.03$. (It is left as an exercise for the reader to verify these solutions.) The time is $t=3.96\text{s}$ or $–1.03\text{s}$. The negative value of time implies an event before the start of motion, and so we discard it. Thus,

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

From the information now in hand, we can find the final horizontal and vertical velocities $v_x$ and $v_y$ and combine them to find the total velocity $v$ and the angle ${\theta }_0$ it makes with the horizontal. Of course, $v_x$ is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore:

The final vertical velocity is given by the following equation:

where $v_{\mathrm{0y}}$ was found in part (a) to be $\text{14}\text{.}\text{3 m/s}$. Thus,

so that

To find the magnitude of the final velocity $v$ we combine its perpendicular components, using the following equation:

which gives

The direction ${\theta }_v$ is found from the equation:

so that

Thus,

The negative angle means that the velocity is $\text{50}\text{.}\mathrm{1º}$ below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 3.39.)