To determine when the two blocks will be at the same height again, we need to consider the time it takes for each block to reach the same vertical displacement.
Let's assume the initial height of both blocks is h, and they start simultaneously. We'll denote the time it takes for block 1 to reach the same height again as t₁, and the time for block 2 as t₂.
For block 1: The inclined plane makes an angle of 50 degrees with the horizontal. The acceleration along the incline can be determined using the formula: a₁ = g * sin(50°)
For block 2: The inclined plane makes an angle of 40 degrees with the horizontal. The acceleration along the incline can be determined using the formula: a₂ = g * sin(40°)
Since block 2 starts with an initial velocity of 2 m/s, its position as a function of time is given by: s₂ = 2t₂ + (1/2) * a₂ * t₂²
For both blocks, the initial position (h) is the same. We need to find the time when the positions of the blocks are the same again, so: s₁ = s₂
Setting up the equation: h + (1/2) * a₁ * t₁² = 2t₂ + (1/2) * a₂ * t₂²
We also know that the vertical displacements of the blocks are equal: h = (1/2) * a₂ * t₂²
Substituting this into the equation: (1/2) * a₂ * t₂² + (1/2) * a₁ * t₁² = 2t₂ + (1/2) * a₂ * t₂²
Simplifying the equation: (1/2) * a₁ * t₁² = 2t₂
Substituting the expressions for a₁ and a₂: (1/2) * g * sin(50°) * t₁² = 2t₂
We also know that the time taken by block 2, t₂, can be related to t₁ using the velocity equation: 2t₂ = 2 / sin(40°) * t₁
Substituting this into the equation: (1/2) * g * sin(50°) * t₁² = 2 / sin(40°) * t₁
Simplifying the equation: g * sin(50°) * t₁² = 4 / sin(40°) * t₁
Dividing both sides by t₁: g * sin(50°) * t₁ = 4 / sin(40°)
Simplifying further: t₁ = (4 / sin(40°)) / (g * sin(50°))
Using the value of the acceleration due to gravity, g ≈ 9.8 m/s², we can evaluate the right side of the equation and find the value of t₁.
Note: It's important to make sure the angles are in radians when using the trigonometric functions.