2.9 Vertical motion under gravity (application)

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.

Representative problems and their solutions

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 :

• Motion plots
• Equal displacement
• Equal time
• Displacement in a particular second
• Twice in a position
• Collision in air

Motion plots

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 :

$$\begin{array}{l}{v}^2=0+2gh\\ \Rightarrow v=\surd \left(2gh\right)=\surd (2x10x80)=40m/s\end{array}$$

Corresponding speed is :

$$\begin{array}{l}\left|v\right|=40m/s\end{array}$$

The time to reach the floor is :

$$\begin{array}{l}t=\frac{v-u}{a}=\frac{40-0}{10}=4s\end{array}$$

According to question, the ball moves up with half the speed. Hence, its speed after first strike is :

$$\begin{array}{l}\left|v\right|=20m/s\end{array}$$

The corresponding upward velocity after first strike is :

$$\begin{array}{l}v=-20m/s\end{array}$$

It reaches a maximum height, when its speed and velocity both are equal to zero. The time to reach maximum height is :

$$\begin{array}{l}t=\frac{v-u}{a}=\frac{0+20}{10}=2s\end{array}$$

After reaching the maximum height, the ball returns towards floor and hits it with the same speed with which it was projected up,

$$\begin{array}{l}v=20m/s\end{array}$$

Corresponding speed is :

$$\begin{array}{l}\left|v\right|=20m/s\end{array}$$

The time to reach the floor is :

$$\begin{array}{l}t=2s\end{array}$$

Again, the ball moves up with half the speed with which ball strikes the floor. Hence, its speed after second strike is :

$$\begin{array}{l}\left|v\right|=10m/s\end{array}$$

The corresponding upward velocity after first strike is :

$$\begin{array}{l}v=-10m/s\end{array}$$

It reaches a maximum height, when its speed and velocity both are equal to zero. The time to reach maximum height is :

$$\begin{array}{l}t=\frac{v-u}{a}=\frac{0+10}{10}=1s\end{array}$$