A helicopter is ascending vertically with a speed of $8.0 m s^{- 1}$ . At a height of $12 m$ above the earth, a package is dropped from its window. How much time does it taken for the package to reach the ground?

1.23 s

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3.23 s

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5.83 s

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2.53 s

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Solution

The correct option is D $2.53 s$
R.E.F. Image.
U = 8 cm/sec.,a = -g
s $= u t + \frac{1}{2} a t^{2}$
$- 12 = 8 t - \frac{1}{2} \times 9 t^{2}$
$- 12 = 8 t - 5 t^{2}$
$5 t^{2} - 8 t - 12 = 0$
t = 2.53 seconds.
so, option (d)

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A helicopter is ascending vertically with a speed of 8.0ms−1. At a height of 12m above the earth, a package is dropped from its window. How much time does it taken for the package to reach the ground?

The correct option is D 2.53sR.E.F. Image.U = 8 cm/sec.,a = -gs =ut+12at2−12=8t−12×9t2−12=8t−5t25t2−8t−12=0t = 2.53 seconds.so, option (d)

A helicopter is ascending vertically with a speed of $8.0 m s^{- 1}$ . At a height of $12 m$ above the earth, a package is dropped from its window. How much time does it taken for the package to reach the ground?

The correct option is D $2.53 s$
R.E.F. Image.
U = 8 cm/sec.,a = -g
s $= u t + \frac{1}{2} a t^{2}$
$- 12 = 8 t - \frac{1}{2} \times 9 t^{2}$
$- 12 = 8 t - 5 t^{2}$
$5 t^{2} - 8 t - 12 = 0$
t = 2.53 seconds.
so, option (d)