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An object falls freely from rest near the surface of Earth. What is the speed of the object after having fallen a distance at 4.90 meters?
(1) 4.90 m/s
(2) 9.80 m/s
(3) 24.0 m/s
(4) 96.1 m/s

Respuesta :

v^2 = vi^2  + 2as

vi = 0 in this case
a = 9.8 m/s^2

So,

V^2 =  2 * 9.8 * 490
v = 98.05 m/s

hope this helps

aachen

Answer:

v = 9.8 m/s

Explanation:

It is given that,

Distance covered by the object, s = 4.9 m

Initial speed of the object, u = 0 (at rest)

It is moving under the action of gravity such that a = g. Using the equation of kinematics as :

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

[tex]v=\sqrt{2gs}[/tex]

[tex]v=\sqrt{2\times 9.8\times 4.9}[/tex]

v = 9.8 m/s

So, the speed of the object after having fallen is 9.8 m/s. Hence, this is the required solution.

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