Answer:
Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x– and y-axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: {a}_{y}=-g=-9.80 m{\text{/s}}^{2}. (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, {a}_{x}=0. Both accelerations are constant, so the kinematic equations can be used.
e-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.
Solution for (a)
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
y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\text{gt}}^{2}\text{.}
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}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{0} = (\text{25}\text{.}\text{0 m/s})(\text{sin 35.0º}) = \text{14}\text{.}\text{3 m/s}. Substituting known values yields
-\text{20}\text{.}0 m\text{}=\left(\text{14}\text{.}3 m/s\text{}\right)t-\left(4\text{.}\text{90 m/s}{\text{}}^{2}\right){t}^{2}\text{.}
Rearranging terms gives a quadratic equation in t:
\left(4\text{.}\text{90 m/s}{\text{}}^{2}\right){t}^{2}-\left(\text{14}\text{.}\text{3 m/s}\right)t-\left(\text{20.0 m}\right)=0.
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=-20.0. Its solutions are given by the quadratic formula:
t=\frac{-b±\sqrt{{b}^{2}-4\text{ac}}}{\text{2}\text{a}}\text{.}
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\phantom{\rule{0.25em}{0ex}}\text{s} or -1.03\phantom{\rule{0.25em}{0ex}}\text{s}. The negative value of time implies an event before the start of motion, and so we discard it. Thus,
t=3\text{.}\text{96 s}\text{.}
Discussion for (a)
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.
Solution for (b)
https://phet.colorado.edu/sims/projectile-motion/projectile-motion_en.html
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