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In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. Part A If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?'

Respuesta :

Answer:

[tex]v_y = 7.76 m/s[/tex]

Explanation:

As we know that the helicopter is ascending up with certain speed so when the person release his briefcase then it will also move up with the same speed.

So here we know that

[tex]\Delta y = v_y t + \frac{1}{2} at^2[/tex]

so here we have

[tex]\Delta y = -130 m[/tex]

[tex]a = -9.81 m/s^2[/tex]

[tex]t = 6 s[/tex]

now we have

[tex]-130 = v_y(6) + \frac{1}{2}(-9.81)(6^2)[/tex]

[tex]v_y = 7.76 m/s[/tex]

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