A tire swing is released from some initial height such that the speed of the tire at the
bottom of the swing is 2.5 m/s. What is the initial height of the tire? Note: g = 9.8 m/s2
A 0.13m
B 0.16m
C 0.32m
D 3.1 m

The Initial height of the tire will be 0.32 m

Explanation:

We  have the formula,

h=Vf^2/2g

h=V×f×f/2g

where Vf denotes the speed of the tire at the bottom of the swing

where g denotes the acceleration

due to the gravity

h=2.5×2.5=6.25

2×9.8 =19.8

h=6.25/19.8

h=0.32

The Initial height of the tire will be 0.32 m

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.

no sign of dia of fuel cell