A balloon is ascending at the rate $v = 12 k m / h$ and is being carried horizontally by the wind at $v_{w} = 20 k m / h .$ If a ballast bag is dropped from the balloon at the instant $h = 50 m$ , determine the time needed for it to strike the ground. Assume that the bag was released from the balloon with the same velocity as the balloon. Also, find the speed with which bag the strikes the ground?

3.55 s, 32.7 m / s

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4.55 s, 40 m / s

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3.55 s, 40 m / s

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4.55 s, 32.7 m / s

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Solution

The correct option is A $3.55 s, 32.7 m / s$
$u = 12 k m p h = 3.33 m / s s = u t - \frac{1}{2} g t^{2} \Rightarrow - 50 = 3.33 t - 5 t^{2} \Rightarrow t = 3.55 s v_{y} = u_{y} - g t = 3.33 - 10 \times 3.55 = - 32.17 m / s v_{x} = 20 k m p h = 5.55 m / s v = \sqrt{{v_{x}}^{2} + {v_{y}}^{2}} = \sqrt{{5.55}^{2} + {32.17}^{2}} = 32.7 m / s$

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The correct option is A 3.55s,32.7m/su=12kmph=3.33m/ss=ut−12gt2⇒−50=3.33t−5t2⇒t=3.55svy=uy−gt=3.33−10×3.55=−32.17m/svx=20kmph=5.55m/sv=√vx2+vy2=√5.552+32.172=32.7m/s