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(B) $1$. Initial motion of the balloon: The balloon starts from rest $(u = 0)$ with acceleration $(a = 2 \, m/s^2)$. After $t = 1 \, s$,the velocity o

Vedclass · Accepted Answer

$0.7$

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(B) $1$. Initial motion of the balloon: The balloon starts from rest $(u = 0)$ with acceleration $(a = 2 \, m/s^2)$. After $t = 1 \, s$,the velocity of the balloon is $v = u + at = 0 + 2(1) = 2 \, m/s$. The height reached by the balloon is $h = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2)(1)^2 = 1 \, m$.$2$. Motion of the stone: When the stone is dropped,it has an initial velocity equal to the balloon's velocity at that instant,which is $u_s = 2 \, m/s$ (upwards). The stone is at a height of $1 \, m$ and is subject to gravity ($g = 9.8 \, m/s^2$ downwards).$3$. Using the equation of motion for the stone: $s = u_s t - \frac{1}{2}gt^2$. Here,$s = -1 \, m$ (since it falls to the ground below the starting point).$-1 = 2t - \frac{1}{2}(9.8)t^2$$-1 = 2t - 4.9t^2$$4.9t^2 - 2t - 1 = 0$$4$. Solving the quadratic equation using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$t = \frac{2 \pm \sqrt{(-2)^2 - 4(4.9)(-1)}}{2(4.9)} = \frac{2 \pm \sqrt{4 + 19.6}}{9.8} = \frac{2 \pm \sqrt{23.6}}{9.8}$Taking the positive root: $t = \frac{2 + 4.858}{9.8} \approx \frac{6.858}{9.8} \approx 0.7 \, s$.

$A$ balloon starts rising from the ground from rest with an upward acceleration of $2 \, m/s^2$. Just after $1 \, s$,a stone is dropped from it. The time taken by the stone to strike the ground is nearly ....... $s$.

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