2.8 Vertical motion under gravity  (Page 4/5)

1: In order to describe motion of the stone once it is released, we realize that it would be easier for us if we shift the origin to the point where stone is released. Considering origin at the point of release and upward direction as positive as shown in the figure, the displacement during the motion of stone is :

y = OB = - 40 m

2: Distance, on the other hand, is equal to :

s = OA + AO + OB = 2 OA + OB

In order to obtain, OA, we consider this part of rectilinear motion (origin at the point of release and upward direction as positive as shown in the figure).

Here, u = 10 m/s; a = -10 $m/s^2$ and v = 0. Applying equation of motion, we have :

$$\begin{array}{l}{v}^2=u^2+2ay\\ y=\frac{v^2-u^2}{2a}=\frac{0^2-{10}^2}{-2x10}=5m\end{array}$$

Hence, OA = 5 m, and distance is :

s = 2 OA + OB = 2 x 5 + 40 = 50 m

3: The time of the journey of stone after its release from the balloon is obtained using equation of motion (origin at the point of release and upward direction as positive as shown in the figure).

Here, u = 12 m/s; a = -10 $m/s^2$ and y = - 40 m.

$$\begin{array}{l}y=ut+\frac{1}{2}a{t}^2\\ \Rightarrow -40=10t-0.5x10t^2\\ \Rightarrow 5t^2-10t-40=0\end{array}$$

This is a quadratic equation in “t”. Its solution is :

$$\begin{array}{l}\Rightarrow 5t^2+10t-20t-40=0\\ \Rightarrow 5t(t+2)-20(t+2)=0\\ \Rightarrow (5t-20)(t+2)=0\\ \Rightarrow 4,-2s\end{array}$$

As negative value of time is not acceptable, time to reach the ground is 4s.

Position

We use the equation $\Delta x=x_2-x_1=ut+\frac{1}{2}a{t}^2$ normally in the context of displacement, even though the equation is also designed to determine initial ( $x_1$ ) or final position ( $x_2$ ). In certain situations, however, using this equation to determine position rather than displacement provides more elegant adaptability to the situation.

Let us consider a typical problem highlighting this aspect of the equation of motion.