Search any question & find its solution
Question:
Answered & Verified by Expert
Two stones are thrown up simultaneously from the edge of a cliff $200 \mathrm{~m}$ high with initial speeds of $15 \mathrm{~ms}^{-1}$ and 30 $\mathrm{ms}^{-1}$. Verify that the graph shown in Fig. correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$. Give the equations for the linear and curved parts of the plot.
Solution:
1010 Upvotes
Verified Answer
For first stone, $x(0)=200 \mathrm{~m}, v(0)=15 \mathrm{~ms}^{-1}$, $a=-10 \mathrm{~ms}^{-2}$
$$
\begin{aligned}
&x_1(t)=x(0)+v(0) t+\frac{1}{2} a t^2 \\
&x_1(t)=200+15 t-5 t^2
\end{aligned}
$$
When the first stone hits the ground, $x_1(t)=0$
$$
\therefore \quad-5 t^2+15 t+200=0
$$
On simplification, $t=8 \mathrm{~s}$
For second stone, $x(0)=200 \mathrm{~m}, \mathrm{v}(0)$
$$
\begin{aligned}
=30 \mathrm{~ms}^{-1}, \mathrm{a} &=-10 \mathrm{~ms}^{-2} \\
x_2(t)=200+30 t-5 t^2
\end{aligned}
$$
When this stone hits the ground, $x_2(t)=0$
$$
\therefore \quad-5 t^2+30 t+200=0
$$
Relative position of second stone w.r.t. first is given by $x_2(t)-x_1(t)=15 t$
Since there is a linear relationship between $x_2(t)-x_1(t)$ and $t$, therefore the graph is a straight line.
For maximum separation, $t=8 \mathrm{~s}$
So maximum separation is $120 \mathrm{~m}$
After 8 second, only the second stone would be in motion.
So, the graph is in accordance with the quadratic equation.
$$
\begin{aligned}
&x_1(t)=x(0)+v(0) t+\frac{1}{2} a t^2 \\
&x_1(t)=200+15 t-5 t^2
\end{aligned}
$$
When the first stone hits the ground, $x_1(t)=0$
$$
\therefore \quad-5 t^2+15 t+200=0
$$
On simplification, $t=8 \mathrm{~s}$
For second stone, $x(0)=200 \mathrm{~m}, \mathrm{v}(0)$
$$
\begin{aligned}
=30 \mathrm{~ms}^{-1}, \mathrm{a} &=-10 \mathrm{~ms}^{-2} \\
x_2(t)=200+30 t-5 t^2
\end{aligned}
$$
When this stone hits the ground, $x_2(t)=0$
$$
\therefore \quad-5 t^2+30 t+200=0
$$
Relative position of second stone w.r.t. first is given by $x_2(t)-x_1(t)=15 t$
Since there is a linear relationship between $x_2(t)-x_1(t)$ and $t$, therefore the graph is a straight line.
For maximum separation, $t=8 \mathrm{~s}$
So maximum separation is $120 \mathrm{~m}$
After 8 second, only the second stone would be in motion.
So, the graph is in accordance with the quadratic equation.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.