An object falls freely from rest near the surface of earth. What is the speed of the object when it has fallen 4.9 meters from its rest position ?

1 Answer
Mar 20, 2018

9.8 m/s

Explanation:

Listed below are the four kinematic equations. Keep in mind that the kinematic equations are the same for the x-direction and y-direction for one-dimensional motion, except that the variable x is replaced with the variable y, in the equations:
http://motionfreefall.blogspot.com/p/kinematic-equations-for-free-fall.htmlhttp://motionfreefall.blogspot.com/p/kinematic-equations-for-free-fall.html

We're going to call the downwards direction the positive y-direction, which means as the object falls down, it's position is increasing in the y-direction

The quantities we know are:
- Acceleration, a = +g = 9.8 m/s^2, because the object is falling downwards, and acceleration caused by gravity causes the object's speed to increase
- Displacement Δy: 4.9 m
- Initial Speed v_i: 0 m/s

Because we are looking for the final speed, v_f, and we don't know the time, choose the equation that excludes t, and solve for v_f. In this case, it would be the third equation, and we would pick the third equation, where v_f = sqrt(v_i^2 + 2*a*Δy). Therefore, after plugging numbers, v_f = 9.8 m/s (which happens to also be the value for acceleration caused by gravity).

Hope this helps!