- #1

Santilopez10

- 81

- 8

- Homework Statement
- A boy decides to hang its bag from a tree branch, with the purpose of raising it. The boy walks with constant velocity in the x axis, and the bag moves in the y axis only. Taking into account that the lenght of the rope is constant and that we know the height h of the branch respect to the floor:

a) find the acceleration of the bag

- Relevant Equations
- kinematic equations

So what I did was at first consider the case the kid is below the branch, so that x=0,t=0, then I thought that the length L of the rope should be ##L=2h## because we know the radius from the branch to the kid is just ##x^2+y^2=r^2## and when x=0, y=h. So then I wrote the motion equations for the bag: $$a(t)=a-g$$ $$x(t)=\frac{(a-g)t^2}{2}$$

And I want to find the time t so that the bag is at the branch, then x(t)=h: $$h=\frac{(a-g)t^2}{2} \rightarrow \sqrt{\frac{2h}{a-g}}=t$$

At this moment, ##x=v_0 \sqrt{\frac{2h}{a-g}}## then we have that the length of the rope is ##h^2+(v_0 \sqrt{\frac{2h}{a-g}})^2=4h^2## and we find that $$a= \frac{2{v_0}^2}{3h}+g$$

Is this correct?

And I want to find the time t so that the bag is at the branch, then x(t)=h: $$h=\frac{(a-g)t^2}{2} \rightarrow \sqrt{\frac{2h}{a-g}}=t$$

At this moment, ##x=v_0 \sqrt{\frac{2h}{a-g}}## then we have that the length of the rope is ##h^2+(v_0 \sqrt{\frac{2h}{a-g}})^2=4h^2## and we find that $$a= \frac{2{v_0}^2}{3h}+g$$

Is this correct?