One-dimensional motion with constant acceleration

[tja2468]

Homework Statement

A falling object travels one-fourth of its total distance in the last second of its fall. From what height was it dropped?

Homework Equations

equations of motion for constant acceleration



Thanks in advance !

 

[Hootenanny]

Homework Statement

A falling object travels one-fourth of its total distance in the last second of its fall. From what height was it dropped?

Homework Equations

equations of motion for constant acceleration
Thanks in advance !

You seem to have omitted a vital section,

The Attempt at a Solution

 

[tomcue]

Based on the information given, we can use the equations of motion for constant acceleration to solve for the initial height of the falling object. We know that the distance traveled in the last second is one-fourth of the total distance, which means that in the last second, the object covered three-fourths of the total distance. This can be represented by the equation d = ut + 1/2at^2, where d is the total distance, u is the initial velocity (which we can assume is 0), a is the acceleration, and t is the time.

So, plugging in the values, we get 3/4d = 1/2at^2. Now, we also know that the total time of the fall is equal to the time taken to cover the last one-fourth of the distance, which is 1 second. Therefore, we can write the equation as d = 1/2at^2, where t=1 second.

Solving for a, we get a = 2d. Now, we can use this value of a in the equation v = u + at, where v is the final velocity (which we can assume is 0 since the object hits the ground), u is the initial velocity (again, assumed to be 0), and a is the acceleration we just calculated.

This gives us 0 = 0 + 2d * 1 second, which simplifies to d = 0. This means that the object was dropped from a height of 0, which makes sense since it is falling towards the ground. Therefore, the initial height of the object was 0 meters.

Hope this helps!