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a helicopter is ascending vertically. a passenger accidentally drops her wallet out the sides of the helicopter when it is 160 m above the ground. if the wallet hits the ground 7s later, at what speed was the helicopter ascending at the moment that the passenger let go of the wallet?

(A) 11.4 m/s
(B) 7.7 m/s
(C) 57.2 m/s
(D) 68.6 m/s
(E) 56.0 m/s

Respuesta :

lublana

Answer:

(E)56.0 m/s

Explanation:

Height =h=-160 m

Because the wallet moving in downward direction

Time=t=7 s

Final speed of wallet=v=0

We have to find the speed of helicopter ascending  at the moment when the passenger let go of the wallet.

[tex]v^2-u^2=2gh[/tex]

Where [tex]g=9.8 m/s^2[/tex]

Substitute the values

[tex]0-u^2=2(-160)\times 9.8[/tex]

[tex]u^2=3136[/tex]

[tex]u=\sqrt{3136}=56m/s[/tex]

Option (E) is true

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