Find the distance travelled by the particle in 0-4 s and 4-6 s.
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Answer:
Solution for (b)
As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use y={y}_{0}+\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t. Because {y}_{0} is zero, this equation reduces to simply
y=\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t\text{.}
Note that the final vertical velocity, {v}_{y}, at the highest point is zero. Thus,
\begin{array}{lll}t& =& \frac{2y}{\left({v}_{0y}+{v}_{y}\right)}=\frac{2\left(\text{233 m}\right)}{\left(\text{67.6 m/s}\right)}\\ & =& \text{6.90 s}\text{.}\end{array}
Discussion for (b)
How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed {v}_{0}, the greater the range, as shown in (Figure)(a). The initial angle {\theta }_{0} also has a dramatic effect on the range, as illustrated in (Figure)(b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with {\theta }_{0}=\text{45º}. This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately \text{38º}. Interestingly, for every initial angle except \text{45º}, there are two angles that give the same range—the sum of those angles is \text{90º}. The range also depends on the value of the acceleration of gravity g. The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range R of a projectile on level ground for which air resistance is negligible is given by
R=\frac{{v}_{0}^{2}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{2\theta }_{0}}{g}\text{,}
where {v}_{0} is the initial speed and {\theta }_{0} is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described.
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