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A helicopter accelerates vertically from the ground from rest at 2 m/s2. 3 s after the helicopter leaves the ground a mailbag is dropped from the helicopter. What is the speed of the mailbag just before it hits the ground? (The helicopter is moving when the bag is dropped.)

Respuesta :

Answer:

The velocity of the mailbag just before it hits the ground is 14.57 m/s.

Explanation:

Given that,

Acceleration = 2 m/s

Time = 3 sec

We need to calculate the velocity of mailbag

Using equation of motion

[tex]v=u+at[/tex]

Put the value into the formula

[tex]v=0+2\times3[/tex]

[tex]v = 6 m/s[/tex]

We need to calculate the height at which the mailbag dropped

Using equation of motion

[tex]H=ut+\dfrac{1}{2}at^2[/tex]

Put the value into the formula

[tex]H=0+\dfrac{1}{2}\times2\times(3)^2[/tex]

[tex]H=9\ m[/tex]

We need to calculate the velocity of the mailbag just before it hits the ground

Using equation of motion

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

[tex]v=\sqrt{u^2+2gh}[/tex]

Put the value into the formula

[tex]v=\sqrt{6^2+2\times9.8\times9}[/tex]

[tex]v=14.57\ m/s[/tex]

Hence, The velocity of the mailbag just before it hits the ground is 14.57 m/s.

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