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A balloon is ascending vertically at a rate v=12 km/h and its horizontal speed is $20 km / h$ . If a ballast bag is dropped from the balloon at the instant h=50 m, find the speed with which the bag strikes the ground. Assume that the bag was released from the balloon with the same velocity as the balloon.

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30.7 m/s

32.3 m/s

34.7 m/s

36.7 m/s

answer is B.

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Detailed Solution

$u_{x} = 20 km / h = 20 \times \frac{5}{18} = 5.6 m / s u_{y} = 12 km / h = 12 \times \frac{5}{18} = 3.3 m / s$

Using $s = u t + \frac{1}{2} a t^{2}$ in vertical direction, we have

$- 50 = (3.3) t + \frac{1}{2} (- 10) t^{2}$

Solving this equation, we get

t = 3.51 s

At the time of striking with ground,

$v_{x} = u_{x} = 5.6 m / s v_{y} = u_{y} + a_{y} t = (3.3) + (- 10) (3.51) = - 31.8 m / s$

Speed $= \sqrt{(- 31.8)^{2} + (5.6)^{2}}$

$\approx 32.3 m / s$

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