Find the initial velocity ##U##

[chwala]

Homework Statement

see attached.

Relevant Equations

Mechanics

Now in determining the initial velocity;

in my understanding, if $s=1.8$ then we consider the stone's motion from the top to the ground. Why not consider $s=3.6$, the total distance traveled by stone from start point $t=0$? Is it possible to model equations from this point?

The stone when thrown upwards will reach a point where it is instantaneously at rest and then start the descent. In that case, it is clear that $s=1.8$. I need insight on this very part.

 

[kuruman]

Why not consider $s=3.6$, the total distance traveled by stone from point $t=0$?

Because in the kinematic equations the relevant variable is the displacement (a vector) not the distance (a scalar).

 

[chwala]

Thank @kuruman I am self-studying Mechanics. Noted.

 

[kuruman]

Thank @kuruman I am self-studying Mechanics. Noted.

Then I will add here a calculation that illustrates the point in post #2.

Problem
A stone is thrown straight up in the air with initial velocity $v_0=3~$m/s and returns to the point at which it was launched. Find the time time of flight $T$.

Solution
The height of the stone $y$ at any time $t$ above the point of launch is given by $$y=v_0~t-\frac{1}{2}g~t^2.$$ At the specific time $t=T$ when it returns to the launch point, the height of the stone above ground is zero. With $t=T$ the equation we have $$ 0=v_0~T-\frac{1}{2}g~T^2\implies T(v_0 -\frac{1}{2}g~T)=0.$$ One or the other of the terms in the product on the right must be zero.

A. $T=0$ which says that the stone is at zero height when it is launched, a fact that we already knew and built in the equation. We reject this solution because we are looking for a time after launch, i.e. $T>0.$

B. $v_0 -\frac{1}{2}g~T=0 \implies T=\dfrac{2v_0}{g}.$ This is the solution that we want.

Answer
$T=\dfrac{2\times 3~(\text{m/s})}{10~(\text{m/s}^2)}=0.6~\text{s}.$

You can see from this example that the total distance traveled up and down does not enter the picture and is not needed. Good luck with your self-study. We are here to help.