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Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.
1: Identify projectile motion types. The possible variants are :
2: We can not use standard equations of time of flight, maximum height and horizontal range. We need to analyze the problem in vertical direction for time of flight and maximum height. Remember that determination of horizontal range will involve analysis in both vertical (for time of flight) and horizontal (for the horizontal range) directions.
3: However, if problem has information about motion in horizontal direction, then it is always advantageous to analyze motion in horizontal direction.
We discuss problems, which highlight certain aspects of the study leading to the projectile motion types. The questions are categorized in terms of the characterizing features of the subject matter :
Problem : A ball from a tower of height 30 m is projected down at an angle of 30° from the horizontal with a speed of 10 m/s. How long does ball take to reach the ground? (consider g = 10 $\phantom{\rule{2pt}{0ex}}m/{s}^{2}$ )
Solution : Here, we consider a reference system whose origin coincides with the point of projection. Further, we consider that the downward direction is positive y - direction.
Motion in vertical direction :
Here, $\phantom{\rule{2pt}{0ex}}{u}_{y}=u\mathrm{sin}\theta =10\mathrm{sin}{30}^{0}=5\phantom{\rule{2pt}{0ex}}m/s;y=30\phantom{\rule{2pt}{0ex}}m;$ . Using $\phantom{\rule{2pt}{0ex}}y={u}_{y}t+\frac{1}{2}{a}_{y}{t}^{2}$ , we have :
$$\begin{array}{l}\Rightarrow \mathrm{30}=5t+\frac{1}{2}10{t}^{2}\\ \Rightarrow {t}^{2}+t-6=0\\ \Rightarrow t(t+3)-2(t+3)=0\\ \Rightarrow t=-3\phantom{\rule{2pt}{0ex}}s\phantom{\rule{2pt}{0ex}}\mathrm{or}\phantom{\rule{2pt}{0ex}}t=2\phantom{\rule{2pt}{0ex}}s\end{array}$$
Neglecting negative value of time, t = 2 s
Problem : A ball is thrown from a tower of height “h” in the horizontal direction at a speed “u”. Find the horizontal range of the projectile.
Solution : Here, we consider a reference system whose origin coincides with the point of projection. we consider that the downward direction is positive y - direction.
$$\begin{array}{l}\Rightarrow x=R={u}_{x}T=uT\end{array}$$
Motion in the vertical direction :
Here, $\phantom{\rule{2pt}{0ex}}{u}_{y}=0\phantom{\rule{2pt}{0ex}}$ and t = T (total time of flight)
$$\begin{array}{l}\Rightarrow h=\frac{1}{2}g{T}^{2}\\ \Rightarrow T=\sqrt{\left(\frac{2h}{g}\right)}\end{array}$$
Putting expression of total time of flight in the expression for horizontal range, we have :
$$\begin{array}{l}\Rightarrow R=u\sqrt{\left(\frac{2h}{g}\right)}\end{array}$$
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