From a balloon moving upwards with a velocity | vedclass

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(A) The initial velocity of the packet is the same as the velocity of the balloon,so $u = 12\,ms^{-1}$.Since the packet is released from a height,its

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$5$

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(A) The initial velocity of the packet is the same as the velocity of the balloon,so $u = 12\,ms^{-1}$.Since the packet is released from a height,its displacement when it reaches the ground is $s = -65\,m$ (taking the upward direction as positive).The acceleration due to gravity is $g = -10\,ms^{-2}$.Using the equation of motion $s = ut + \frac{1}{2}at^2$:$-65 = 12t + \frac{1}{2}(-10)t^2$$-65 = 12t - 5t^2$$5t^2 - 12t - 65 = 0$Solving the quadratic equation using the formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$t = \frac{12 \pm \sqrt{(-12)^2 - 4(5)(-65)}}{2(5)}$$t = \frac{12 \pm \sqrt{144 + 1300}}{10}$$t = \frac{12 \pm \sqrt{1444}}{10}$$t = \frac{12 \pm 38}{10}$Since time cannot be negative,we take $t = \frac{12 + 38}{10} = \frac{50}{10} = 5\,s$.

From a balloon moving upwards with a velocity of $12\,ms^{-1}$,a packet is released when it is at a height of $65\,m$ from the ground. The time taken by it to reach the ground is...........$s$ $(g = 10\,ms^{-2})$.

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