2.8 Vertical motion under gravity  (Page 2/5)

$$\begin{array}{l}\mathrm{For}v=-u,\\ v=u+at=u-gt\\ \Rightarrow -u=u-gt\\ \Rightarrow t=\frac{2u}{g}\end{array}$$

The time taken for the complete journey is twice the time taken to reach the maximum height. It means that the ball takes equal time in upward and downward journey. Thus, the total motion can be considered to be divided in two parts of equal duration.

5: The velocity of the ball is positive in the first half of motion; Zero at the maximum height; negative in the second of the motion.

6: The velocity is decreasing all through the motion from a positive value to less positive value in the first half and from a less negative value to more negative value in the second half of the motion. This renders acceleration to be always negative (directed in -y direction), which is actually the case.

7: The velocity (positive) and acceleration (negative) in the first part are opposite in direction and the resulting speed is decreasing. On the other hand, the velocity (negative) and acceleration (negative) in the second part are in the same direction and the resulting speed is increasing.

Displacement and distance

Let us analyze the equation $\Delta x=x_2-x_1=ut+\frac{1}{2}a{t}^2$ for the vertical motion under gravity with the help of earlier example. If we choose initial position as the origin, then $x_1=0,$ $\Delta x=x_2=x\left(\mathrm{say}\right)$ and $x=ut+\frac{1}{2}a{t}^2$ , where x denotes position and displacement as well. In the frame of reference with upward direction as positive,

$$\begin{array}{l}u=30m/s\mathrm{and}a=-g=-10m/s^2\end{array}$$

Putting these values in the equation, we have :

$$\begin{array}{l}\Rightarrow x=30t-5t^2\end{array}$$

The important aspect of this equation is that it is a quadratic equation in time “t”. This equation yields two values of time “t” for every position and displacement. This outcome is in complete agreement with the actual motion as the ball reaches a given position twice (during upward and downward motion). Only exception is point at the maximum height, which is reached only once. We have seen earlier that ball reaches maximum height at t = 3 s. Therefore, maximum height,H, is given as :

$$\begin{array}{l}\Rightarrow H=30X3-5X9=45m\end{array}$$

The displacement values for the motion at successive seconds are :

------------------------------------------------------------- Time (t) ut 5txt Displacementin seconds or position (x) in meters------------------------------------------------------------- 0.0 0 0 01.0 30 5 25 2.0 60 20 403.0 90 45 45 4.0 120 80 405.0 150 125 25 6.0 180 180 0-------------------------------------------------------------

The corresponding displacement – time plot looks like as shown in the figure.

We notice following important characteristics of the motion :

1: The ball retraces every position during motion except the point at maximum height.

2: The net displacement when ball return to initial position is zero. Thus, the total time of journey (T) is obtained using displacement, x = 0,

$$\begin{array}{l}\mathrm{For}x=0,\\ x=ut+\frac{1}{2}a{t}^2=uT-\frac{1}{2}g{t}^2=0\\ \Rightarrow 2uT-g{T}^2=0\\ \Rightarrow T=\frac{\mathrm{2u}}{g}\end{array}$$

Here, we neglect T = 0, which corresponds to initial position.

3: The “x” in equation $x=ut+\frac{1}{2}a{t}^2$ denotes displacement and not distance. Hence, it is not possible to use this equation directly to obtain distance, when motion is not unidirectional.