A balloon is rising vertically with a velocity of $9.8$ $m / s$ . A packet is dropped from it when it is at a height of $39.2$ m. Time taken by the packet to reach the ground is (Given $g = 9.8 m / s^{2}$ )

1 s

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2 s

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3 s

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4 s

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Solution

The correct option is D $4 s$
Packet acquire the ballon velocity after it is dropped
$- 39.2 = 9.8 t - \frac{1}{2} g t^{2}$
$\Rightarrow 5 t^{2} - 9.8 t - 39.2 = 0$
$t = \frac{9.8 \pm \sqrt{(9.8)^{2} + 4 \times 39.2 \times 5}}{10} = 3.94 s e c$ $\approx 4 s e c$

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A balloon is rising vertically with a velocity of 9.8 m/s. A packet is dropped from it when it is at a height of 39.2 m. Time taken by the packet to reach the ground is (Given g=9.8m/s2)

The correct option is D 4sPacket acquire the ballon velocity after it is dropped−39.2=9.8t−12gt2⇒5t2−9.8t−39.2=0t=9.8±√(9.8)2+4×39.2×510=3.94 sec ≈4 sec

A balloon is rising vertically with a velocity of $9.8$ $m / s$ . A packet is dropped from it when it is at a height of $39.2$ m. Time taken by the packet to reach the ground is (Given $g = 9.8 m / s^{2}$ )

The correct option is D $4 s$
Packet acquire the ballon velocity after it is dropped
$- 39.2 = 9.8 t - \frac{1}{2} g t^{2}$
$\Rightarrow 5 t^{2} - 9.8 t - 39.2 = 0$
$t = \frac{9.8 \pm \sqrt{(9.8)^{2} + 4 \times 39.2 \times 5}}{10} = 3.94 s e c$ $\approx 4 s e c$