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  • A ballon starts ascending at a constant acceleration of <img alt="2 \mathrm{~m} / \mathrm{s}^2" src="https://entrancecorner.oncodecogs.com/gif.latex?2%20%5Cmathrm%7B%7Em%7D%20/%20%5Cmathrm%7Bs%7D%5

A ballon starts ascending at a constant acceleration of 2 \mathrm{~m} / \mathrm{s}^2. When it was at a height of 100 \mathrm{~m} from the ground, the food packet is dropped from the ballon. After how much time and with what velocity does it reach the ground?

Option: 1

49\ m/s


Option: 2

40\ m/s


Option: 3

56 \ m/s


Option: 4

4\ m/s


Answers (1)

best_answer

The velocity of ballon at the height of 100 \mathrm{~m}
\begin{aligned} & v^2=u^2+2 a s=0+2 \times 2 \times 100 \\ & v^2=400 \\ & v=\sqrt{400}=20 \mathrm{~m} / \mathrm{s} . \end{aligned}
Take this velocity -20 \mathrm{~m} / \mathrm{s} along downward direction The food packet is falling with initial velocity -20\ \mathrm{m/s} from a height 100 \mathrm{~m}
Therefore from second equation,
\begin{aligned} & h=u t+\frac{1}{2} g t^2 \\ & 100=-20 t+\frac{1}{2} \times 10 \times t^2 \\\ & 100=-20 t+5 t^2 \\\ & 5 t^2-20 t-100=0 \\\ & t^2-4 t-20=0 \end{aligned}               
solving fort, we get t=2+\sqrt{24} \mathrm{sec}
The velocity with with it strikes the ground -
By Using formula ,  \frac{-b_{-}^{+}\sqrt{b^{3}+4ac}}{2a}
\begin{aligned} & v^2=u^2+2 g h \\ & v^2=(-20)^2+2 \times 10 \times 100 \\ & v^2=400+2000=2400 \\ & k^2=2400=24 \times 100 \\ & v=\sqrt{24 \times 100}=10 \sqrt{24} \end{aligned}
v = 49\ m/s

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Rishi

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