2.8 Vertical motion under gravity  (Page 3/5)

Let us answer the question with respect to the motion of the ball under consideration : what is the distance traveled in first 4 seconds? Obviously, the ball travels 30 m in the upward direction to reach maximum height in 3 seconds and then it travels 5 m in the $4^{\mathrm{th}}$ second in downward direction. Hence, the total distance traveled is 45 + 5 = 50 m in 4 s. This means that we need to apply the equation of motion in two parts : one for the upward motion and the second for the downward motion. Thus, we find displacement for each segment of the motion and then we can add their magnitude to obtain distance.

The distance values for the motion at successive seconds are :

------------------------------------------------------------ Time (t) ut 5t*t Displacement Distancein seconds or position in meters (x) in meters------------------------------------------------------------ 0.0 0 0 0 01.0 30 5 25 25 2.0 60 20 40 403.0 90 45 45 45 4.0 120 80 40 505.0 150 125 25 65 6.0 180 180 0 90------------------------------------------------------------

Constant acceleration

Problem : A balloon starts rising from the ground with an acceleration of 1.25 m/s. After 8 second, a stone is released from the balloon. Starting from the release of stone, find the displacement and distance traveled by the stone on reaching the ground. Also, find the time taken to reach the ground (take g = 10 $m/s^2$ ).

Solution : This question raises few important issues. First the rise of balloon is at a constant acceleration of 1.25 $m/s^2$ . This acceleration is the “measured” acceleration, which is net of the downward acceleration due to gravity. This means that the balloon rises with this net vertical acceleration of 1.25 $m/s^2$ in the upward direction.

Here, u = 0; a = 1.25 $m/s^2$ ; t = 8 s. Let the balloon rises to a height “h” during this time, then (considering origin on ground and upward direction as positive) the displacement of the balloon after 8 seconds is :

$$\begin{array}{l}& y=ut+\frac{1}{2}a{t}^2=0x8+0.5x1.25x{8}^2=40m\end{array}$$

Now, we know that, the body released from moving body acquires the velocity but not the acceleration of the container body under motion. The velocity of the balloon at the instant of separation is equal to the velocity of the balloon at that instant.

$$\begin{array}{l}& v=u+at=0+1.25x8=10m/s\end{array}$$

Thus, this is the initial velocity of the stone and is directed upward as that of the velocity of balloon. Once released, the stone is acted upon by the force of gravity alone. The role of the acceleration of the balloon is over. Now, the acceleration for the motion of stone is equal to the acceleration due to gravity, g.

The path of motion of the stone is depicted in the figure. Stone rises due to its initial upward velocity to a certain height above 40 m where it was released till its velocity is zero. From this highest vertical point, the stone falls freely under gravity and hits the ground.