A balloon is cut at a height of 81m. and going up with velocity of 12m/s. If a block is releaed from it, then in how much time it will reach the ground?

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Solution

Given,
$u = 12 m / s$
$h = 81 m$
$g = 10 m / s^{2}$
from kinematics, $h = u t - \frac{1}{2} g t^{2}$
$81 = 12 t - \frac{1}{2} \times 10 t^{2}$
$- 5 t^{2} + 12 t - 81 = 0$ (quadratic equation)
It's solution is,
$t_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}, t_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}$
$t_{1} = \frac{- 12 + \sqrt{12^{2} - 4 \times (- 5) \times 81}}{(- 10)}$
$t_{1} = \frac{- 12 + 42}{- 10} = \frac{30}{10} = - 3$ (don't take negative value)
$t_{2} = \frac{- 12 - \sqrt{12^{2} - 2 \times (- 5) \times 81}}{(- 10)}$
$t_{2} = \frac{- 12 - 42}{- 10} = \frac{54}{10}$
$t_{2} = 5.4 s e c$

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