Question

A helicopter is ascending with a velocity of 2 m/s at a height of 24 m when it drops a mail pocket. The packet contains material, which can be damaged if it hits ground with velocity of magnitude greater than 72 km/h. Was the packet damaged? Explain your answer. [Take g= 10 m/s²)​

Answer 1

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$\huge\bold\purple{Answer:-}$

⏩ Here, u= -2 m/s, s= -24 m, g= -10 m/s²

v² = u² + 2gs = (-2)² + 2 × (-24) × (-10)

= 4 + 480 = 484

or v= ± √484

or v= -22 m/s [Downward velocity is taken as " -ve " ]

Now,

→Safe velocity = 72 km/h

= 72 × 1000/3600 = 20 m/s

→ As the packet reaches the ground with a velocity (22 m/s) greater than the safe velocity of 20 m/s, so it is damaged.

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Hope it helps...:-)

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Answer 2

Explanation:

Here, u= -2 m/s, s= -24 m, g= -10 m/s²

v² = u² + 2gs = (-2)² + 2 × (-24) × (-10)

= 4 + 480 = 484

or v= ± √484

or v= -22 m/s [Downward velocity is taken as " -ve " ]

→Safe velocity = 72 km/h

= 72 × 1000/3600 = 20 m/s

→ As the packet reaches the ground with a velocity (22 m/s) greater than the safe velocity of 20 m/s, so it is damaged.

✔✔✔✔✔✔