g you slide a toy car to see the loop the loop physics. the radius of the loop 0.5 m. The acceleration of gravity what is the minimal height h you need to release the car from so that it makes the loop

 h = 1.5 m

Explanation:

This exercise must be solved using the concepts of energy.

Starting point. Place where the car leaves

          Em₀ = U = m g h

Final point. Highest part of the loop, r = 2R

          Em_f = K + U

          Em_f = ½ m v² + mg (2R)

the energy is preserved

          Em₀ = Em_f

          m g h = ½ m v² + m g 2R               (1)

the car must have a minimum velicate so that it does not fall, let's use Newton's second law.

         ∑ F = m a

         N + W = m a

acceleration is centripetal

          a = v²/ r

the normal is the reaction of the surface to the support of the car, as the speed decreases, the normal decreases until reaching zero

         0 + m g = m v² / r

         v² =  rg

we substitute

        v² = 2Rg

we substitute in equation 1

          g h = ½ 2Rg + 2Rg

          h = 3R

let's calculate

          h = 3 0.5

          h = 1.5 m

50 inches should be the resultant vector