Question

39. A balloon is moving up with const velocity 20 m/s. At
height 20 m a packet is released. Find time to reach
of ground after the packet is released.
(1) 2+2/2 (2) 2 + √2
(3) 2-√2 (4) 2√2-2
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Answer 1

Answer:

When the packet is release it's initial velocity will be 20 m /s upward equal to balloon due to its inertia .

Use energy conservation

$\frac{1}{2} m {u}^{2} + mgh = \frac{1}{2} m {v}^{2} \\ \\ {v}^{2} = {u}^{2} +2 gh \\ {v}^{2} = {20}^{2} + 2 \times 10 \times 20 \\ {v}^{2} = 400 + 400 \\ v = 20 \sqrt{2}$

Time

Let upward direction positive .

u = +20 m/s

v = - 20√2

a = -g = -10 m/s2

so

$t = \frac{v - u}{g} = \frac{ - 20 \sqrt{2} - 20}{ - 10} = 2 \sqrt{2} + {2}$