From a balloon moving upwards with a velocity of $12 m s^{- 1}$ , 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
(take, $g = 10 m s^{- 2})$

5 s

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8 s

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4 s

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7 s

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10 s

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Solution

The correct option is A $5 s$
Given, $u = 12 m s^{- 1}$
$s = - 65 m$
$\Rightarrow g = - 10 m s^{- 2}$
We know that, $s = u t + \frac{1}{2} g t^{2}$
$\Rightarrow - 65 = 12 t - \frac{1}{2} \times 10 t^{2}$
$- 65 = 12 t - 5 t^{2}$
$\Rightarrow 5 t^{2} - 12 t - 65 = 0$
$5 t^{2} - 25 t + 13 t - 65 = 0$
$5 t (t - 5) + 13 (t - 5) = 0$
$(5 t + 13) (t - 5) = 0$
$\Rightarrow t = - \frac{13}{5} s$ or $t = 5 s$ .

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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(take, g=10 ms−2)

The correct option is A 5 sGiven, u=12 ms−1s=−65 m⇒g=−10 ms−2We know that, s=ut+12gt2⇒−65=12t−12×10t2−65=12t−5t2⇒5t2−12t−65=05t2−25t+13t−65=05t(t−5)+13(t−5)=0(5t+13)(t−5)=0⇒t=−135s or t=5s.

From a balloon moving upwards with a velocity of $12 m s^{- 1}$ , 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
(take, $g = 10 m s^{- 2})$