# 5.3 Features of projectile motion (application)

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Solving problems is an essential part of the understanding process.

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.

## Hints on solving problems

1: In general, we should rely on analysis in two individual directions as linear motion.

2: Wherever possible, we should use the formula directly as available for time of flight, maximum height and horizontal range.

3: We should be aware that time of flight and maximum height are two attributes of projectile motion, which are obtained by analyzing motion in vertical direction. For determining time of flight, the vertical displacement is zero; whereas for determining maximum height, vertical component of velocity is zero.

4: However, if problem has information about motion in horizontal direction, then it is always advantageous to analyze motion in horizontal direction. It is so because motion in horizontal direction is uniform motion and analysis in this direction is simpler.

5: The situation, involving quadratic equations, may have three possibilities : (i) quadratic in time "t" (ii) quadratic in displacement or position "x" and (iii) quadratic in "tanθ" i.e."θ". We should use appropriate equations in each case as discussed in the module titled " Features of projectile motion ".

## Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to the features of projectile motion. The questions are categorized in terms of the characterizing features of the subject matter :

• Time of flight
• Horizontal range
• Maximum height
• Height attained by a projectile
• Composition of motion
• Projectile motion with wind/drag force

## Time of flight

Problem : The speed of a particle, projected at 60°, is 20 m/s at the time of projection. Find the time interval for projectile to loose half its initial speed.

Solution : Here, we see that final and initial speeds (not velocity) are subject to given condition. We need to use the given condition with appropriate expressions of speeds for two instants.

$u=\sqrt{\left({u}_{x}^{2}+{u}_{y}^{2}\right)}$

$v=\sqrt{\left({v}_{x}^{2}+{v}_{y}^{2}\right)}$

According to question,

$u=2v$

$⇒{u}^{2}=4{v}^{2}$

$⇒{u}_{x}^{2}+{u}_{y}^{2}=4\left({v}_{x}^{2}+{v}_{y}^{2}\right)$

But, horizontal component of velocity remains same. Hence,

$⇒{u}_{x}^{2}+{u}_{y}^{2}=4\left({u}_{x}^{2}+{v}_{y}^{2}\right)$

Rearranging for vertical velocity :

${v}_{y}^{2}={u}_{y}^{2}-3{u}_{x}^{2}={\left(20\mathrm{sin}{60}^{0}\right)}^{2}-3X{\left(20\mathrm{cos}{60}^{0}\right)}^{2}$

${v}_{y}^{2}=20X\frac{3}{4}-3X20X\frac{1}{4}=0$

The component of velocity in vertical direction becomes zero for the given condition. This means that the projectile has actually reached the maximum height for the given condition. The time to reach maximum height is half of the time of flight :

$t=\frac{u\mathrm{sin}\theta }{g}=\frac{20X\mathrm{sin}{60}^{0}}{10}=\sqrt{3}\phantom{\rule{1em}{0ex}}s$

## Horizontal range

Problem : A man can throw a ball to a greatest height denoted by "h". Find the greatest horizontal distance that he can throw the ball (consider g = 10 $\phantom{\rule{2pt}{0ex}}m/{s}^{2}$ ).

Solution : The first part of the question provides the information about the initial speed. We know that projectile achieves greatest height in vertical throw. Let "u" be the initial speed. We can, now, apply equation of motion " $\phantom{\rule{4pt}{0ex}}{v}^{2}={u}^{2}+2gy\phantom{\rule{4pt}{0ex}}$ " for vertical throw. We use this form of equation as we want to relate initial speed with the greatest height.

Here, v = 0; a = -g

$\begin{array}{l}⇒0={u}^{2}-2gh\\ {u}^{2}=2gh\end{array}$

The projectile, on the other hand, attains greatest horizontal distance for the angle of projection, θ = 45°. Accordingly, the greatest horizontal distance is :

$\begin{array}{l}{R}_{\mathrm{max}}=\frac{{u}^{2}\mathrm{sin}2X\mathrm{45°}}{g}=\frac{{u}^{2}\mathrm{sin}\mathrm{90°}}{g}=\frac{{u}^{2}}{g}=\frac{2gh}{g}=2h\end{array}$

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