Chapter 2 - Motion Along a Straight Line - Problems - Page 35: 54b

The required graph with both plots is given below :

The first stone is dropped. Thus, the magnitude of initial velocity $u$ is zero. That is, $u=0$ $m/s$. We have also given that the height $s$ for both stones is $43.9$ $m$. That is, $s=43.9$ $m$. Since the second stone was thrown $1$ $s$ after the first is dropped, and both stones reached the water surface at the same time. The second stone takes $1$ $s$ less than the first stone to reach the water surface. That is, if the time taken for the first stone is $t$, then the time taken by the second stone is $t-1$. Here, acceleration $a$ is acceleration due to gravity $g$ and the magnitude of $g$ is $9.81$ $m/s^2$. Using the equation of motion $s=ut+\dfrac{1}{2}at^2$ for first stone. We get, $43.9=0\times t+\dfrac{1}{2}\times9.81t^2$ $\implies 43.9=\dfrac{9.81}{2}t^2$ $\implies t^2=\dfrac{2\times43.9}{9.81}$ $\implies t=\sqrt{\dfrac{2\times43.9}{9.81}}$ , time is only positive $\implies t\approx 2.99166$ $s$ Thus, the time interval for the first stone is $ t\approx 2.99166$ $s$. And, the time interval for the second stone is $t-1\approx 1.99166$ $s$. Now, using the equation of motion $s=ut+\dfrac{1}{2}at^2$ for the second stone. We get, $43.9=u\times (1.99166)+\dfrac{1}{2}\times9.81(1.99166)^2$ $\implies 43.9=u\times (1.99166)+19.4567$ $\implies u\times (1.99166)=43.9-19.4567$ $\implies u\times (1.99166)=24.4433$ $\implies u=\dfrac{24.4433}{1.99166}$ $\implies u\approx 12.2727$ $m/s$ Thus, the magnitude of the initial velocity for the second stone is approximately $12.2727$ $m/s$. Now, using the equation of motion $v=u+at$ for the first stone. We get, $v=0+9.81t\implies v=9.81t$. Now, using the equation of motion $v=u+at$ for the second stone. We get, $v=12.2727+9.81(t-1)$. Plot the equation $v=9.81t$ for first stone from time $0$ $s$ to $2.99166$ $s$. Take $t=0$ $s$, We get, $v=0$ $m/s$ Take $t=2.99166$ $s$, We get, $v=29.34822$ $m/s$ Thus, the end points of the plot are $(0,0)$ and $(2.99166,29.34822)$. Make a line joining these end points to get a plot of velocity versus time of the first stone. Plot the equation $v=12.2727+9.81(t-1)$ for second stone from time $1$ $s$ to $2.99166$ $s$. Take $t=1$ $s$, We get, $v=12.2727$ $m/s$ Take $t=2.99166$ $s$, We get, $v=31.81099$ $m/s$ Thus, the end points of the plot are $(1,12.2727)$ and $(2.99166,31.81099)$. Make a line joining these end points to get a plot of velocity versus time of the second stone. Thus, we get the following graph with both plots.