From a balloon moving upwards with a velocity | vedclass

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Question

$s=u t+\frac{1}{2} g t^{2}$
$\Rightarrow-65=12 t-\frac{1}{2} 	imes 10 t^{2}$
$-65=12 t-5 t^{2}$
$\Rightarrow 5 t^{2}-12 t-65=0$
$5 t^{2}-25 t+13

Vedclass · Accepted Answer

$5$

Vedclass · Answer

$s=u t+\frac{1}{2} g t^{2}$
$\Rightarrow-65=12 t-\frac{1}{2} 	imes 10 t^{2}$
$-65=12 t-5 t^{2}$
$\Rightarrow 5 t^{2}-12 t-65=0$
$5 t^{2}-25 t+13 t-65=0$
$5 t(t-5)+13(t-5)=0$
$(5 t+13)(t-5)=0$
$\Rightarrow t=-\frac{13}{5} s$ or $t=5 s$

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...........$s$ $(g = 10\,ms^{-1})$,

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