Question

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

Answer 1

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 :

$v^2-u^2=2gs$

$v=\sqrt{2gs}$

$v=\sqrt{2\times 9.8\times 4.9}$

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.

Answer 2

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