A stone is dropped from a balloon going up with a uniform velocity of $5 m / s$ . If the balloon was $60 m$ high when the stone was dropped, find the height of the balloon when the stone hits the ground. Take $g = 10 m /$ $s^{2}$ .

100 m

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80 m

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20 m

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50 m

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Solution

The correct option is D $80 m$

$s = u t + \frac{1}{2} a t^{2}$

$- 60 = 5 (t) + \frac{1}{2} (- 10) t^{2}$

$- 60 = 5 t - 5 t^{2}$

$5 t^{2} - 5 t - 60 = 0$

$t^{2} - t - 12 = 0$

$t^{2} - 4 t + 3 t - 12 = 0$

$(t - 4) (t + 3) = 0$

$∴ t = 4$

Height of balloon from ground at this instant

$= 60 + 4 \times 5 = 80 m$

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