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There are classic situations relating to the projectile motion, which needs to be handled with appropriate analysis. We have quite a few ways to deal with a particular situation. It is actually the nature of problem that would determine a specific approach from the following :
Besides, we may require combination of approached as listed above. In this section, we shall study these classic situations involving projectile motion.
A projectile can clear posts of equal height, as projectile retraces vertical displacement attained during upward flight while going down. We can approach such situation in two alternative ways. The equation of motion for displacement yields two values for time for a given vertical displacement (height) : one corresponds to the time for upward flight and other for the downward flight as shown in the figure below. Corresponding to these two time values, we determine two values of horizontal displacement (x).
Alternatively, we may use equation of trajectory of the projectile. The y coordinate has a quadratic equation in "x". it again gives two values of "x" for every value of "y".
Problem : A projectile is thrown with a velocity of $25\sqrt{2}$ m/s and at an angle 45° with the horizontal. The projectile just clears two posts of height 30 m each. Find (i) the position of throw on the ground from the posts and (ii) separation between the posts.
Solution : Here, we first use the equation of displacement for the given height in the vertical direction to find the values of time when projectile reaches the specified height. The equation of displacement in vertical direction (y) under constant acceleration is a quadratic equation in time (t). Its solution yields two values for time. Once two time instants are known, we apply the equation of motion for uniform motion in horizontal direction to determine the horizontal distances as required. Here,
$$\begin{array}{l}{u}_{y}=25\sqrt{2}X\mathrm{sin}{45}^{0}=25\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
$$\begin{array}{l}y={u}_{y}t-\frac{1}{2}g{t}^{2}\\ \Rightarrow \mathrm{30}=\mathrm{25}t-\frac{1}{2}X\mathrm{10}X{t}^{2}\\ \Rightarrow {t}^{2}-5t+6=0\\ \Rightarrow t=2\phantom{\rule{2pt}{0ex}}s\phantom{\rule{2pt}{0ex}}\mathrm{or}\phantom{\rule{2pt}{0ex}}3\phantom{\rule{2pt}{0ex}}s\end{array}$$
Horizontal motion (refer the figure) :
$$\begin{array}{l}\mathrm{OA}={u}_{x}t==25\sqrt{2}x\mathrm{cos}{45}^{0}X2=50\phantom{\rule{2pt}{0ex}}m\end{array}$$
Thus, projectile needs to be thrown from a position 50 m from the pole. Now,
$$\begin{array}{l}\mathrm{OB}={u}_{x}t==25\sqrt{2}x\mathrm{cos}{45}^{0}X3=75\phantom{\rule{2pt}{0ex}}m\end{array}$$
Hence, separation, d, is :
$$\begin{array}{l}d=\mathrm{OB}-\mathrm{OA}=75-50=25\phantom{\rule{2pt}{0ex}}m\end{array}$$
Alternatively
Equation of vertical displacement (y) is a quadratic equation in horizontal displacement (x). Solution of equation yields two values of "x" corresponding to two positions having same elevation. Now, equation of projectile is given by :
$$\begin{array}{l}y=x\mathrm{tan}\theta -\frac{g{x}^{2}}{2{u}^{2}{\mathrm{cos}}^{2}\theta}\end{array}$$
Putting values,
$$\begin{array}{l}\mathrm{30}=x\mathrm{tan}\mathrm{45\xb0}-\frac{10{x}^{2}}{2{\sqrt{\mathrm{25}}}^{2}{\mathrm{cos}}^{2}\mathrm{45\xb0}}\\ \Rightarrow \mathrm{30}=x-\frac{10{x}^{2}}{2{25\sqrt{2}}^{2}{\mathrm{cos}}^{2}\mathrm{45\xb0}}\\ \Rightarrow \mathrm{30}=x-\frac{{x}^{2}}{125}\\ \Rightarrow {x}^{2}-\mathrm{125}x+\mathrm{3750}=0\\ \Rightarrow (x-\mathrm{50})(x-\mathrm{75})=0\\ \Rightarrow x=\mathrm{50}\phantom{\rule{2pt}{0ex}}\mathrm{or}\phantom{\rule{2pt}{0ex}}\mathrm{75}\phantom{\rule{2pt}{0ex}}m\\ \Rightarrow d=\mathrm{75}-\mathrm{50}=\mathrm{25}\phantom{\rule{2pt}{0ex}}m\end{array}$$
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