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A cargo helicopter, descending steadily at a speed of 2.3 m/s, releases a small package. Let upward be the positive direction for this problem.
(a) If the package is 71 m above the ground when it is dropped, how long does it take for the package to reach the ground?
1 s

(b) What is its velocity just before it lands? (Indicate the direction with the sign of your answer.)
2 m/s

Respuesta :

Explanation:

Given that,

Given that,

Speed of the helicopter, [tex]u_x=2.3\ m/s[/tex]

Vertical velocity of the helicopter, [tex]u_y=0[/tex]

(a) Height of the package, h = 71 m

We need to find the time taken for the package to reach the ground. It can be calculated using second equation of motion as :

[tex]h=u_yt+\dfrac{1}{2}at^2[/tex]

Here a = g

[tex]h=\dfrac{1}{2}gt^2[/tex]

[tex]t=\sqrt{\dfrac{2h}{g}}[/tex]

[tex]t=\sqrt{\dfrac{2\times 71}{9.8}}[/tex]

t = 3.80 seconds

(b) We can find the vertical velocity of package. It is given by :

[tex]v_y=u_y-gt[/tex]

[tex]v_y=-gt[/tex]

[tex]v_y=-9.8\times 3.8=-37.24\ m/s[/tex]

Let v is the velocity of the package just before it lands. It is given by :

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]

[tex]v=\sqrt{2.3^2+37.24^2}[/tex]

v = 37.31 m/s

Hence, this is the required solution.

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