Two objects are thrown simultaneously from the same height at 45° angles to the vertical with a

speed of 20 m per second; one up, the other one down. Find the difference between the heights the
objects will be at two seconds later. How are these objects moving in regards to one another?

Y1 = 20 sin45*t - (g/2)t^2

Y2 = -20 sin45*t -(g/2)t^2

Y1 - Y2 = 40 sin45*t = 56.6 m apart

V1 = 20 sin45 - gt
V2 = -20 sin45 - gt

The vertical velocity difference between them remains
V2 - V1 = -40 sin 45 = -28.3 m/s

Two seconds after release, both are going down, but V2 is going down faster.

To find the difference between the heights the objects will be at two seconds later, we first need to determine the height reached by each object after two seconds.

Given that both objects are thrown with a speed of 20 m/s at a 45° angle to the vertical, we can break down the initial velocity into its horizontal and vertical components.

The initial velocity (v) can be split into its horizontal component (vₓ) and vertical component (vᵧ) using trigonometry:

vₓ = v * cos(45°)
vᵧ = v * sin(45°)

Substituting the given speed value:

vₓ = 20 m/s * cos(45°) ≈ 14.14 m/s
vᵧ = 20 m/s * sin(45°) ≈ 14.14 m/s

Since the objects are thrown simultaneously from the same height, their initial height (h) is the same for both.

Now, let's calculate the heights each object will reach after two seconds.

The vertical distance traveled by each object after two seconds (sᵧ) can be calculated using the equation:

sᵧ = vᵧ * t + (1/2) * g * t²

Where:
vᵧ = initial vertical velocity = 14.14 m/s
t = time elapsed = 2 s
g = acceleration due to gravity = 9.8 m/s²

Plugging in the values:

sᵧ = 14.14 m/s * 2 s + (1/2) * 9.8 m/s² * (2 s)²
= 28.28 m + 19.6 m
≈ 47.88 m

Since one object is thrown upwards and the other downwards, the height reached by the object thrown upwards will be positive, while the height reached by the object thrown downwards will be negative. Therefore, the difference between their heights after two seconds will be the sum of these heights:

Difference in height = sᵧ (upwards) + sᵧ (downwards)
= 47.88 m + (-47.88 m)
= 0 m

So, the objects will be at the same height after two seconds.

Regarding how these objects are moving in relation to each other, one is moving upwards while the other is moving downwards. They are moving in opposite directions.