Question

A Helicopter ascending with a Velocity of 2m/s at a height 24cm when it's drop a mail packet . The packet contains material , which can be damaged if it's hit ground with a velocity greater than 72 km/h . Was the packet damaged , Explain and Solve the whole Question .

Answer 1

Heya.....!!!!

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-- Given in the Question :-

• Velocity during ascent = 2 m/s
• Height of the helicopter = 24 m
• height from packet is dropped = 24 m
• initial velocity of packet = - 2m/s

• Final velocity of packet = ??

- Using the equations of motion , we get

=> v^2 - u^2 = 2×a×s

=> v^2 - ( - 2)^2 = 2 × 10 × 24

=> v^2 = 480 + 4 => 484

=> v = √484 = 22 m/s

-----> as the limit of velocity is given 72 km/h the unit is Km/h so ,,,

=> Converting the units of velocity we get ,,

=> v = 22 × 18/5

♦ finally v ➡➡ 79.2 Km/h

Hence ,, the packet hits the ground with velocity of 79.2 Km/h

➡ It will get Damaged , because the safety limit is given to us is 72 km/h .

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Hope It Helps You ^_^

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

$\boxed{\begin{minipage}{11 cm} Given \\ \\ \ Velocity=2 m/s \\ \\ Height=24m \\ \\ Height\;of\;packet\;dropped = 24m \\ \\ u=-2m/s \\ \\ To\;Find:- \\ \\ v = ? \end{minipage}}$

$\textbf{\boxed{Third\; Equation\;of\;Motion}}$

${\boxed{v^{2} = u^{2} + 2as}}$

${\boxed{We\;may\;also\;write\;it\;as}}$

${\boxed{v^{2}-u^{2} = 2as}}$

${\boxed{Substitute\;the\;values}}$

${\boxed{v^{2}-(-2)^{2} = 2\times 10\times 24}}$

= 480 m²/s²

${\boxed{v^{2}=480+4}}$

${\boxed{v^{2}=484}}$

${\boxed{v=\sqrt{484}}}$

= 22 m/s

${\boxed{Now\;here\;we\;have\;to\;covert\;the\;unit}}$

$\tt{\rightarrow v=\dfrac{22\times 1m}{1s}$

$\tt{\rightarrow v=\dfrac{22\times 10^{-3}km}{(1/60\times 60)h}$

${\boxed{22\times 10^{-3}\times 60\times 60km/h}}$

= 79.2 km/h

★Here we get that packet hits ground with high velocity so the limit is given in question that for safety it required 72km/h

★So

${\boxed{Packet\;will\;be\;damaged}}$