3.26 Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s–1 and 30 m s–1. Verify that the graph shown in Fig. 3.27 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 g = 10 m s–2. Give the equations for the linear and curved parts of the plot.
Equation (3) represents the linear trajectory of the two stones because to this linear relation between (s2 – s1) and t, the projection is straight line till 8s (from the graph)
The maximum distance between the two stones is at t = 8s. So
(s2 – s1) max = 15 x 8 = 120 m. This value has been depicted correctly in the graph.
After 8s, only the second stone is in motion (t for first stone has been found out to be 8s), the trajectory of second stone is obtained in equation (2), = 200 + 30t – 5t2 , which is a curved path.
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<div><div><strong>3.26 </strong>For the first stone, </div></div><p>Initial velocity, u<sub>1</sub> = 15m/s, acceleration, a = -g = -10 m/s</p><p>From the relation s<sub>1</sub>=s<sub>0</sub>+u<sub>1</sub>t+ (1/2)at<sup>2</sup> where</p><p>s<sub>0</sub> = cliff height, s<sub>1</sub> = total height of the fall of the first stone, we get</p><p>s<sub>1 </sub>= 200 + 15t – 5t<sup>2 </sup>………. (1)</p><p>When the stone hit the floor, s<sub>1 </sub>= 0, so the equation (1) becomes</p><p>0 = 200 +15t - 5t<sup>2 </sup>= t<sup>2 </sup>-3t – 40 = (t-8) (t+5) = 0</p><p>So t = 8s or -5s</p><p>Since the stone was thrown at t=0, so t cannot be –ve. Hence t = 8s</p><p>For the second stone, </p><p>Initial velocity, u<sub>1</sub> = 30 m/s, acceleration, a = -g = -10 m/s</p><p>From the relation s<sub>2</sub>=s<sub>0</sub>+u<sub>1</sub>t+ (1/2)at<sup>2</sup> where</p><p>s<sub>0</sub> = cliff height, s<sub>2</sub>= total height of the fall of the second stone, we get</p><p>s<sub>2 </sub>= 200 + 30t – 5t<sup>2 </sup>………. (2)</p><p>When the stone hit the floor, s<sub>2 </sub>= 0, so the equation (2) becomes</p><p>0 = 200 +30t - 5t<sup>2 </sup>= t<sup>2 </sup>-6t – 40 = (t-10) (t+4) = 0</p><p>So t = 10s or -4s</p><p>Since the stone was thrown at t=0, so t cannot be –ve. Hence t = 10s</p><p>Subtracting equation (1) from equation (2), we get</p><p>s<sub>2 – </sub>s<sub>1 </sub>= (200 + 30t – 5t<sup>2 </sup>) – (200 + 15t – 5t<sup>2 </sup>) = 15t ……. (3)</p><p>Equation (3) represents the linear trajectory of the two stones because to this linear relation between (s<sub>2 – </sub>s<sub>1) </sub>and t, the projection is straight line till 8s (from the graph)</p><p>The maximum distance between the two stones is at t = 8s. So</p><p> (s<sub>2 – </sub>s<sub>1</sub>) <sub>max </sub>= 15 x 8 = 120 m. This value has been depicted correctly in the graph.</p><p>After 8s, only the second stone is in motion (t for first stone has been found out to be 8s), the trajectory of second stone is obtained in equation (2), = 200 + 30t – 5t<sup>2 </sup>, which is a curved path.</p>
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