Question

A helicopter is ascending with a velocity of 2 m / s at a height of 24 m when it drops a mail packet . The packet contains material , which can be damaged if it hits the ground with velocity of magnitude greater than 72 km / h . Was the packet damaged ? Explain your answer . ( You can take g = 10 m /s^2 )​

Answer 1

$\large \; {\underline{\underline{\pmb{\orange{\sf{ \; Given \; :- }}}}}}$

• Velocity of Helicopter = 2 m/s
• Height of Helicopter = 24 m
• Magnitude at which it should be thrown = 72 km/h
• Gravity = 10 m/s

$\large \; {\underline{\underline{\pmb{\red{\sf{ \; To \; Find \; :- }}}}}}$

• Packet is damaged or not = ?

$\\ \qquad{\rule{200pt}{2pt}}$

$\large \; {\underline{\underline{\pmb{\purple{\sf{ \; SolutioN \; :- }}}}}}$

$\dag \; {\underline{\underline{\pmb{\sf{ \; Final \; Velocity \; of \; Packet \; :- }}}}}$

$\begin{gathered} \; \; \; \blue\dashrightarrow \; \; \red{\sf{ {v}^{2} = {u}^{2} + 2gh }} \\ \\ \\ \end{gathered}$

$\begin{gathered} \; \; \; \dashrightarrow \; \; \sf{ {v}^{2} = {2}^{2} + 2 \times 10 \times 24 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \; \; \; \dashrightarrow \; \; \sf{ {v}^{2} = 4 + 2 \times 10 \times 24 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \; \; \; \dashrightarrow \; \; \sf{ {v}^{2} = 4 + 2 \times 240 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \; \; \; \dashrightarrow \; \; \sf{ {v}^{2} = 4 + 480 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \; \; \; \dashrightarrow \; \; \sf{ {v}^{2} = 484 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \; \; \; \dashrightarrow \; \; \sf{ v = \sqrt{ \bigg( 484 \bigg) } } \\ \\ \\ \end{gathered}$

$\begin{gathered} \; \; \; \dashrightarrow \; \; {\underline{\boxed{\pmb{\frak{ Final \; Velocity {\small_{(Packet)}} = 22 \; m/s }}}}} \; \green\bigstar \\ \\ \\ \end{gathered}$

$\dag \; {\underline{\underline{\pmb{\sf{ \; Converting \; the \; Velocity \; :- }}}}}$

$\begin{gathered} \qquad \; \implies \; \; \sf { Velocity_{m/s} = 22 \times \dfrac{18}{5} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \implies \; \; \sf { Velocity_{m/s} = \dfrac{396}{5} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \implies \; \; \sf { Velocity_{m/s} = \cancel\dfrac{396}{5} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \implies \; \; {\underline{\boxed{\pmb{\frak{ Velocity_{m/s} = 79.2 \; km/h }}}}} \; \red\bigstar \\ \\ \\ \end{gathered}$

$\therefore \;$ Velocity of Packet is more than 72 km/h .So, the Packet will be damaged .

$\\ \qquad{\rule{200pt}{2pt}}$

$\large \; {\underline{\underline{\pmb{\purple{\sf{ \; Note \; :- }}}}}}$

$\longmapsto$ In the given Equation the values will be as follows :-

• v = Final Velocity
• u = Initial Velocity
• g = Gravity
• h = Height

$\\ \qquad{\rule{200pt}{2pt}}$

Answer 2

$\star\;{\underline{\boxed{\pmb{\green{\frak{\;Given\; :-}}}}}}$

• Initial velocity = 2m/s
• Height of Helicopter = 24 m
• Magnitude at which it should be thrown = 72 km/h
• Value of Gravity = 10 m/s

$\star\;{\underline{\boxed{\pmb{\green{\frak{\;To Find\; :-}}}}}}$

$\huge\rightarrow$ $\color{red}{Was\:the\:packet\: damaged\:=\:?}$

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$\star\;{\underline{\boxed{\pmb{\green{\frak{\; Solution\; :-}}}}}}$

The velocity at which the packet gets damaged

$\huge\rightarrow$ $\tt \: 72 km\:h^-1\:=\:72\:×\:\dfrac{5}{18}$

$\huge\rightarrow$ $\star\;{\underline{\boxed{\pmb{\pink{\frak{\;20ms^-1}}}}}}$

When the packet was dropped from the helicopter, initially it will first move upwards the velocity of 2$ms^-1$ (The velocity of the helicopter since it was inside the helicopter) then come to rest and finally start moving downwards [ Free Fall ]

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$\large\underline\color{orange}{\sf{Case\:1}}$

For the upward motion of the packet for velocity of 2$m^-1$

$\huge\rightarrowtail$ $\boxed{Initial\: velocity\:=\:2ms^-1\:Final\: velocity\:0 \:ms^-1}$

$\huge\rightarrowtail$ Distance covered in upward direction = ?

$\huge\rightarrowtail$ Acceleration due to gravity = 10m$s^-2$

$\color{blue}{Since\:packet\:is\: moving\:away\:from\:earth}$

Using ,

.

$\boxed{v²\:-\:u²\:=\:2gs}$

$\Huge\sf\implies$$\tt \: 0² - 2² = 2 × 10 × s$

$\Huge\sf\implies$$\tt \: s = \dfrac{-4}{-20}$

$\Huge\sf\implies$$\tt \: 0.2m$

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$\large\underline{\sf{Case\:2}}$

For the Downward motion of the packet :

• Initial velocity = 0 m/s
• Final velocity = ?
• Distance of fall = 24 + 0.2 = 24.2 m
• value of Gravity = 10 m$s^-2$

Using ,

.

$\boxed{v²\:-\:u²\:=\:2gs}$

$\huge\rightarrow$ $\tt \: v² - 0² = 2 × 10 × 24.2$

$\Huge\sf\implies$ $\star\;{\underline{\boxed{\pmb{\green{\frak{\;484}}}}}}$

$\Huge\sf\implies$ $\tt \: v = √484$

$\Huge\sf\implies$${\underline{\boxed{\pmb{\pink{\frak{\;22ms^-1}}}}}}$

The packet will hit the ground with the velocity of 22 m$s^-1$ which is more than the maximum value of the packet to avoid the damage (20m$s^-1$ , Therefore it will get damaged .

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