As some molten metal splashes, one droplet flies off to the east with initial speed vi at angle θi above the horizontal, and another droplet flies off to the west with the same speed at the same angle above the horizontal, as inthe figure. In terms of vi and θi, find the distance between the droplets as a function of time. (Use v_i for vi, q_i for θi, and t as appropriate in your equation for the distance.)
Both droplets initially rise and then fall due to gravity. Both droplets have the same altitude at a given time, because they started with the same angle (from horizontal) and speed. Therefore the distance between them is determined by their horizontal coordinate along the east-west axis only. That is proportional to the horizontal component of the initial velocity. The separation is double the distance that either one travels hotizontally, because they go in opposite directions.
The separation is therefore
2 (v_i)cos(q_i)t
To find the distance between the droplets as a function of time, we need to analyze the motion of each droplet separately and then calculate the relative distance between them.
Let's consider the eastward flying droplet first. We can decompose its initial velocity, vi, into horizontal and vertical components as follows:
Velocity in the x-direction: vix = vi*cos(qi)
Velocity in the y-direction: viy = vi*sin(qi)
Similarly, for the westward flying droplet, the components of the initial velocity are:
Velocity in the x-direction: vix = -vi*cos(qi) [Note the negative sign indicating the westward direction]
Velocity in the y-direction: viy = vi*sin(qi)
Now, let's derive an equation for the horizontal position of each droplet as a function of time. The horizontal distance traveled by a droplet can be expressed as:
x = vix * t
For the eastward droplet:
x_east = vix * t = vi*cos(qi)*t
For the westward droplet:
x_west = vix * t = -vi*cos(qi)*t
To find the distance between the droplets as a function of time, we calculate the difference in their horizontal positions:
distance = x_east - x_west
= (vi*cos(qi)*t) - (-vi*cos(qi)*t)
= 2*vi*cos(qi)*t
Therefore, the distance between the droplets as a function of time is given by:
distance = 2*vi*cos(qi)*t, where "vi" represents the initial speed and "qi" represents the angle above the horizontal.
To find the distance between the droplets as a function of time, we need to break down the motion of each droplet into its horizontal and vertical components.
Let's consider the droplet that flies off to the east first. The horizontal component of its velocity remains constant, while the vertical component is affected by gravity. The horizontal component of velocity is given by:
Vx = vi * cos(θi)
The vertical component of velocity is given by:
Vy = vi * sin(θi) - gt
where g is the acceleration due to gravity, and t is time.
Similarly, for the droplet flying off to the west, the horizontal component of velocity is in the opposite direction, so it becomes:
Vx = -vi * cos(θi)
The vertical component of velocity remains the same:
Vy = vi * sin(θi) - gt
Now, to find the distance between the droplets as a function of time, we can consider the horizontal distance traveled by each droplet. Let's call the distance between the droplets as D(t).
The eastward droplet's horizontal motion can be expressed as:
x1 = Vx * t = vi * cos(θi) * t
The westward droplet's horizontal motion can be expressed as:
x2 = Vx * t = -vi * cos(θi) * t
The distance between the droplets is simply the difference between their positions:
D(t) = x1 - x2
= vi * cos(θi) * t - (-vi * cos(θi) * t)
= 2 * vi * cos(θi) * t
Therefore, the distance between the droplets as a function of time is given by:
D(t) = 2 * vi * cos(θi) * t