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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.
We discuss problems, which highlight certain aspects of the study leading to the accelerated motion under gravity. The questions are categorized in terms of the characterizing features of the subject matter :
Problem : A ball is dropped from a height of 80 m. If the ball looses half its speed after each strike with the horizontal floor, draw (i) speed – time and (ii) velocity – time plots for two strikes with the floor. Consider vertical downward direction as positive and g = 10 $m/{s}^{2}$ .
Solution : In order to draw the plot, we need to know the values of speed and velocity against time. However, the ball moves under gravity with a constant acceleration 10 $m/{s}^{2}$ . As such, the speed and velocity between strikes are uniformly increasing or decreasing at constant rate. It means that we need to know end values, when the ball strikes the floor or when it reaches the maximum height.
In the beginning when the ball is released, the initial speed and velocity both are equal to zero. Its velocity, at the time first strike, is obtained from the equation of motion as :
$$\begin{array}{l}{v}^{2}=0+2gh\\ \Rightarrow v=\surd \left(2gh\right)=\surd (2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}10\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}80)=40\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
Corresponding speed is :
$$\begin{array}{l}\left|v\right|=40\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
The time to reach the floor is :
$$\begin{array}{l}t=\frac{v-u}{a}=\frac{40-0}{10}=4\phantom{\rule{2pt}{0ex}}s\end{array}$$
According to question, the ball moves up with half the speed. Hence, its speed after first strike is :
$$\begin{array}{l}\left|v\right|=20\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
The corresponding upward velocity after first strike is :
$$\begin{array}{l}v=-20\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
It reaches a maximum height, when its speed and velocity both are equal to zero. The time to reach maximum height is :
$$\begin{array}{l}t=\frac{v-u}{a}=\frac{0+20}{10}=2\phantom{\rule{2pt}{0ex}}s\end{array}$$
After reaching the maximum height, the ball returns towards floor and hits it with the same speed with which it was projected up,
$$\begin{array}{l}v=20\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
Corresponding speed is :
$$\begin{array}{l}\left|v\right|=20\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
The time to reach the floor is :
$$\begin{array}{l}t=2\phantom{\rule{2pt}{0ex}}s\end{array}$$
Again, the ball moves up with half the speed with which ball strikes the floor. Hence, its speed after second strike is :
$$\begin{array}{l}\left|v\right|=10\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
The corresponding upward velocity after first strike is :
$$\begin{array}{l}v=-10\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
It reaches a maximum height, when its speed and velocity both are equal to zero. The time to reach maximum height is :
$$\begin{array}{l}t=\frac{v-u}{a}=\frac{0+10}{10}=1\phantom{\rule{2pt}{0ex}}s\end{array}$$
Problem : A ball is dropped vertically from a height “h” above the ground. It hits the ground and bounces up vertically to a height “h/3”. Neglecting subsequent motion and air resistance, plot its velocity “v” with the height qualitatively.
Solution : We can proceed to plot first by fixing the origin of coordinate system. Let this be the ground. Let us also assume that vertical upward direction is the positive y-direction of the coordinate system. The ball remains above ground during the motion. Hence, height (y) is always positive. Since we are required to plot velocity .vs. height (displacement), we can use the equation of motion that relates these two quantities :
$$\begin{array}{l}{v}^{2}={u}^{2}+2ay\end{array}$$
During the downward motion, u = 0, a = -g and displacement (y) is positive. Hence,
$$\begin{array}{l}{v}^{2}=-2gy\end{array}$$
This is a quadratic equation. As such the plot is a parabola as motion progresses i.e. as y decreases till it becomes equal to zero (see plot below y-axis).
Similarly, during the upward motion, the ball has certain velocity so that it reaches 1/3 rd of the height. It means that ball has a velocity less than that with which it strikes the ground. However, this velocity of rebound is in the upwards direction i.e positive direction of the coordinate system. In the nutshell, we should start drawing upward motion with a smaller positive velocity. The equation of motion, now, is :
$$\begin{array}{l}{v}^{2}=2gy\end{array}$$
Again, the nature of the plot is a parabola the relation being a quadratic equation. The plot progresses till displacement becomes equal to y/3 (see plot above y-axis). The two plots should look like as given here :
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