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2.Motion in Straight Line
hard
Two stones are thrown up simultaneously from the edge of a cliff $200 \;m$ high with inttial speeds of $15\; m s ^{-1}$ and $30 \;m s ^{-1} .$ Verify that the graph shown in Figure correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect atr resistance and assume that the stones do not rebound after hitting the ground. Take $g=10\; m s ^{-2} .$ Give the equations for the linear and curved parts of the plot.
Option A
Option B
Option C
Option D
Solution
For first stone:
Initial velocity, $u_{1}=15 m / s$
Acceleration, $a=- g =-10 m / s ^{2}$
Using the relation, $x_{1}=x_{0}+u_{1} t+\frac{1}{2} a t^{2}$
Where, height of the cliff, $x_{0}=200 m$ $x_{1}=200+15 t-5 t^{2} … (i)$
When this stone hits the ground, $x_{1}=0$
$\therefore-5 t^{2}+15 t+200=0$
$t^{2}-3 t-40=0$
$t^{2}-8 t+5 t-40=0$
$t(t-8)+5(t-8)=0$
$t=8$ s or $t=-5 s$
since the stone was projected at time $t=0,$ the negative sign before time is meaningless.
$\therefore t=8 s$
For second stone:
Initial velocity, $u_{ II }=30 m / s$
Acceleration, $a=- g =-10 m / s ^{2}$
Using the relation,
$x_{2}=x_{0}+u_{11} t+\frac{1}{2} a t^{2}$
$=200+30 t-5 t^{2}.. .( ii )$
At the moment when this stone hits the ground; $x_{2}=0$
$5 t^{2}+30 t+200=0$
$t^{2}-6 t-40=0$
$t^{2}-10 t+4 t+40=0$
$t(t-10)+4(t-10)=0$
$t(t-10)(t+4)=0$
$t=10 s$ or $t=-4 s$
Here again, the negative sign is meaningless.
$\therefore t=10 s$
Subtracting equations (i) and (ii), we get $x_{2}-x_{1}=\left(200+30 t-5 t^{2}\right)-\left(200+15 t-5 t^{2}\right)$
$x_{2}-x_{1}=15 t …(iii)$
Equation (iii) represents the linear path of both stones. Due to this linear relation between $\left(x_{2}-x_{1}\right)$ and $t,$ the path remains a straight line till 8 s.
Maximum separation between the two stones is at $t=8 s$
$\left(x_{2}-x_{1}\right)_{\max }=15 \times 8=120 m$
This is in accordance with the given graph.
After $8 s$, only second stone is in motion whose variation with time is given by
the quadratic equation: $x_{2}-x_{1}=200+30 t-5 t^{2}$
Hence, the equation of linear and curved path is given by
$x_{2}-x_{1}=15 t \quad$ (Linear path)
$x_{2}-x_{1}=200+30 t-5 t^{2} \quad$ (Curved path)
Initial velocity, $u_{1}=15 m / s$
Acceleration, $a=- g =-10 m / s ^{2}$
Using the relation, $x_{1}=x_{0}+u_{1} t+\frac{1}{2} a t^{2}$
Where, height of the cliff, $x_{0}=200 m$ $x_{1}=200+15 t-5 t^{2} … (i)$
When this stone hits the ground, $x_{1}=0$
$\therefore-5 t^{2}+15 t+200=0$
$t^{2}-3 t-40=0$
$t^{2}-8 t+5 t-40=0$
$t(t-8)+5(t-8)=0$
$t=8$ s or $t=-5 s$
since the stone was projected at time $t=0,$ the negative sign before time is meaningless.
$\therefore t=8 s$
For second stone:
Initial velocity, $u_{ II }=30 m / s$
Acceleration, $a=- g =-10 m / s ^{2}$
Using the relation,
$x_{2}=x_{0}+u_{11} t+\frac{1}{2} a t^{2}$
$=200+30 t-5 t^{2}.. .( ii )$
At the moment when this stone hits the ground; $x_{2}=0$
$5 t^{2}+30 t+200=0$
$t^{2}-6 t-40=0$
$t^{2}-10 t+4 t+40=0$
$t(t-10)+4(t-10)=0$
$t(t-10)(t+4)=0$
$t=10 s$ or $t=-4 s$
Here again, the negative sign is meaningless.
$\therefore t=10 s$
Subtracting equations (i) and (ii), we get $x_{2}-x_{1}=\left(200+30 t-5 t^{2}\right)-\left(200+15 t-5 t^{2}\right)$
$x_{2}-x_{1}=15 t …(iii)$
Equation (iii) represents the linear path of both stones. Due to this linear relation between $\left(x_{2}-x_{1}\right)$ and $t,$ the path remains a straight line till 8 s.
Maximum separation between the two stones is at $t=8 s$
$\left(x_{2}-x_{1}\right)_{\max }=15 \times 8=120 m$
This is in accordance with the given graph.
After $8 s$, only second stone is in motion whose variation with time is given by
the quadratic equation: $x_{2}-x_{1}=200+30 t-5 t^{2}$
Hence, the equation of linear and curved path is given by
$x_{2}-x_{1}=15 t \quad$ (Linear path)
$x_{2}-x_{1}=200+30 t-5 t^{2} \quad$ (Curved path)
Standard 11
Physics
