Question

A particle dropped, if it travel half the distance in last second find time of journey

Answer 1

A body travels half of its total path in the last second of its free fall. What is the duration of its free fall?
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4 ANSWERS
Bhavik Bansal
Bhavik Bansal, High school nerd
Answered Feb 12, 2017 · Author has 76 answers and 354.1k answer views
The first approach to this question is to define the variables. So , let's do that:

Let the total distance from the ground be “h” , so the distance travelled in last one second will be “h/2”.

So as it is free fall , u(initial velocity) = 0

Let's take the velocity of the object after traveling half the distance as “v” and the velocity with which it hits the ground as “v'”

And as always acceleration = 9.8 m/s/s (As you'll see “g” doesn't matter , it cancels out.)

Also time taken to complete first half is taken as “t” and to complete second half is “t'” = 1s.


NOTE : I am gonna take the Equations of motion as they are , if you don't know them please ask Google for derivations.

Now ,

First let's calculate v(velocity after half of the distance has been travelled)

v=u+at(Equationofmotion)v=u+at(Equationofmotion)

(As u = 0 ) the above equation becomes

v=at−−−−−(I)v=at−−−−−(I)

Now , with that out of the way.

Another equation of motion is as you know

S=ut+1/2(at2)(EquationofMotion)S=ut+1/2(at2)(EquationofMotion)

So , as the first and the last half distance travelled is h/2 , we can achieve two more equations specific to our question from the above equation.

h/2=0∗t+1/2(at2h/2=0∗t+1/2(at2)−−−−−−−(II))−−−−−−−(II)(As S = h/2 and u = 0)

Also , for the last second travelled

h/2=ut+1/2(a∗1)−−−−−−−−−(III)h/2=ut+1/2(a∗1)−−−−−−−−−(III) (As t = 1s and 1^2 = 1)

Now remember equation (I) above , yeah I do too ,

Let's replace u of the equation (III) with the v from equation (I) cause initial velocity of the last second travelled is v as defined in the variables.

Now substituting ,

h/2=(at)(t′)+1/2(a)h/2=(at)(t′)+1/2(a) (As v = at)

The Equation becomes ,

h/2=(at)+1/2(a)−−−−−(IV)h/2=(at)+1/2(a)−−−−−(IV) (As t' = 1)

So , now let's equate equation (II) and equation (IV)

(at)+a/2=1/2(at2)(at)+a/2=1/2(at2)

=>at+a/2=(at2)/2=>at+a/2=(at2)/2

=>a(t+1/2)=a(t2)/2=>a(t+1/2)=a(t2)/2

=>t+1/2=t2/2=>t+1/2=t2/2 {As a got cancelled , I told you :)}

=>2t+1=t2=>2t+1=t2 (Taking 2 in the denominator to the other side)

=>t2−2t+1=0