किसी $200 \,m$ ऊँची खड़ी चट्ट|न के किनारे से दो पत्थरों को एक साथ ऊपर की ओर $15\, m\, s ^{-1}$ तथा $30 \,m\, s ^{-1}$ की प्रारंभिक चाल से फेंका जाता है । इसका सत्यापन कीजिए कि नीचे दिखाया गया ग्राफ ( चित्र) पहले पत्थर के सापेक्ष दूसरे पत्थर की आपेक्षिक स्थिति का समय के साथ परिवर्तन को प्रदर्शित करता है । वायु के प्रतिरोध को नगण्य मानिए और यह मानिए कि जमीन से टकराने के बाद् पत्थर ऊपर की ओर उछलते नहीं । मान लिजिए $g=10\, m\, s ^{-2}$ । ग्राफ के रेखीय व वक्रीय भागों के लिए समीकरण लिखिए ।

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)