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A balloonist, riding in the basket of a hot air balloon that is rising vertically with a constant velocity of 11.9 m/s, releases a sandbag when the balloon is 43.4 m above the ground. Neglecting air resistance, what is the bag's speed when it hits the ground?

Respuesta :

Answer:

v = 31.5 m/s

Explanation:

given,

Speed of air balloon, u = 11.9 m/s

height of balloon above the ground, h = 43.4 m

acceleration due to gravity = 9.8 m/s²

Speed of the bag when it hit the ground = ?

using equation of motion

[tex]v^2 = u^2 + 2 g h[/tex]

[tex]v^2 = 11.9^2 + 2\times 9.8\times 43.4[/tex]

[tex]v= \sqrt{992.25}[/tex]

    v = 31.5 m/s

Speed of the bag before it hit the ground is equal to v = 31.5 m/s

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