JEE MAIN - Physics (2015 (Offline) - No. 10)
Two stones are thrown up simultaneously from the edge of a cliff $$240$$ $$m$$ high with initial speed of $$10$$ $$m/s$$ and $$40$$ $$m/s$$ respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ?
(Assume stones do not rebound after hitting the ground and neglect air resistance, take $$g = 10m/{s^2}$$)
(The figures are schematic and not drawn to scale)
(Assume stones do not rebound after hitting the ground and neglect air resistance, take $$g = 10m/{s^2}$$)
(The figures are schematic and not drawn to scale)
_en_10_1.png)
_en_10_2.png)
_en_10_3.png)
_en_10_4.png)
Explanation
Using $$h = ut + {1 \over 2}g{t^2}$$
$${y_1} = 10t - 5{t^2};\,\,{y_2} = 40t - 5{t^2}$$
$$\therefore$$ $$\,\,\,\,{y_2} - {y_1} = 30t\,\,\,\,$$ for $$\,\,\,\,t \le 8s.$$
Curve will be straight line when $$t \le 8s.$$
when stone 1 reaches the ground then $$\,\,\,{y_1} = - 240m,\,\, and\,\, t = 8s$$
for $$\,\,\,\,t > 8s.$$
$${y_2} - {y_1} = 40t + {1 \over 2}g{t^2} - 240$$
So, it will be a parabolic curve till stone 2 reaches the ground. And parabola should opens upward as coefficient of t2 is positive.
$${y_1} = 10t - 5{t^2};\,\,{y_2} = 40t - 5{t^2}$$
$$\therefore$$ $$\,\,\,\,{y_2} - {y_1} = 30t\,\,\,\,$$ for $$\,\,\,\,t \le 8s.$$
Curve will be straight line when $$t \le 8s.$$
when stone 1 reaches the ground then $$\,\,\,{y_1} = - 240m,\,\, and\,\, t = 8s$$
for $$\,\,\,\,t > 8s.$$
$${y_2} - {y_1} = 40t + {1 \over 2}g{t^2} - 240$$
So, it will be a parabolic curve till stone 2 reaches the ground. And parabola should opens upward as coefficient of t2 is positive.
Comments (0)
