Acceleration - The acceleration of a body is defined as the change in its velocity during an interval of time divided by the duration of the time interval. If Vo is the initial velocity at the beginning of the period of time and Vf is the final velocity at the end of the period of time, the change in velocity is Vo - Vf. If the velocity change occurs over the period of time t, the acceleration of the body is given by a = (Vo - Vf)/t.
The relationships between initial velocity, final velocity, distance covered and time, in uniformlay accelerated motion are defined mathematically by the following three equations:
1--From a = (Vo - Vf)/t, the final velocity of a body under constant acceleration is given by Vf = Vo + at.
2--The second equation regarding accelerated motion defines the distance traveled by a body under uniform acceleration.The average velocity of a moving body during a time interval "t" is expressed by Vav = (Vo + Vf)/2. The distance traveled, "s", during this time interval "t" is the product of the average velocity and the duraion of the time interval or s = (Vo + Vf)t/2. Substituting Vf = Vo + at into this expression yields s = [Vo + (Vo + at)]t/2 or s = Vot + at^2/2.
3--The third equation of uniformly accelerated motion is derived from the first two by eliminating the time interval "t". Multiplying the two expressions results in as = (Vf - Vo)/t x (Vo + Vf)t/2 or Vf^2 = Vo^2 + 2as.
In summary,
Vf = Vo + at
s = Vot + at^2/2
Vf^2 = Vo^2 + 2as
These same equations apply to rising and falling bodies with the exception that a is replaced by g, the acceleration due to gravity.
For rising bodys,
Vf = Vo - gt
s = Vot - gt^2/2
Vf^2 = Vo^2 - 2gs.
For falling bodys,
Vf = Vo + gt
s = Vot + gt^2/2
Vf^2 = Vo^2 + 2gs
From s = h = Vot + gt^2/2
Since Vo = 0, h = gt^2 or t = sqrt(2h/g)
Hi
Would someone be able to confirm that the following equation wil give the fall time for an object falling under gravity ignoring air resistance:
T= sq root of 2h/g
where h is fall height and g is accel due to local value for gravity.
A derivation from a more usual equation would be great too.
Thankc
2 answers
Thanks very much
1 answer