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.)
To find the distance between the droplets as a function of time, we need to analyze the horizontal and vertical components of their motion separately.
Let's consider the droplet moving to the east first. The horizontal component of its initial velocity (vi) is given by vi * cos(θi), where θi is the angle above the horizontal. The horizontal distance traveled by this droplet over time can be represented as d1 = (vi * cos(θi)) * t.
Now, let's consider the droplet moving to the west. Since it is moving in the opposite direction, the horizontal component of its initial velocity will be negative, -vi * cos(θi). The horizontal distance traveled by this droplet over time can be represented as d2 = (-vi * cos(θi)) * t.
The distance between the droplets is the sum of these two distances, so we can write it as d = d1 + d2.
Substituting the expressions for d1 and d2, we get:
d = (vi * cos(θi)) * t + (-vi * cos(θi)) * t
Simplifying further, we have:
d = (vi * cos(θi) - vi * cos(θi)) * t
d = 0 * t
d = 0
Therefore, the distance between the droplets as a function of time is always zero. This means that regardless of how much time has passed, the two droplets will always be at the same position.
It's worth noting that this result assumes no external forces acting on the droplets, such as air resistance or gravity.