Daniel B. answered 10/09/22
A retired computer professional to teach math, physics
Let
w = 1.4 m/s be Luna's speed,
v = 2 m/s be Chloe's speed,
α = 13° be the angle of Chloe's jump,
s (to be calculated) be Luna's distance from Chloe at the time of jump,
g = 9.81 m/s² be gravitational acceleration.
Imagine that the two cats are moving in an xy-plane,
where the origin is at the point of Chloe's jump,
the x-axis is horizontal and extends from the point of jump towards Luna,
and y is the vertical axis.
Let xc(t) and yc(t) be Chloe's coordinates at a time t, and
let xl(t) and yl(t) be Luna's coordinates at a time t.
We let time start with the jump, so
xc(0) = 0
yc(0) = 0
xl(0) = s
yl(0) = 0
At any subsequent time t the coordinates are
xc(t) = vcos(α)t
yc(t) = vsin(α)t - gt²/2
xl(t) = s - wt
yl(t) = 0
(If you do not know where these equations come from, let me know.)
We are to find s, so that there exists time t1 when Chloe is on top of Luna,
i.e., their coordinates are the same:
xc(t1) = xl(t1)
yc(t1) = yl(t1)
Substituting xc, yc, xl, yl:
vcos(α)t1 = s - wt1 (1)
vsin(α)t1 - gt1²/2 = 0 (2)
So we have two equations (1), (2) with two unknowns s and t1.
We can rewrite equation (1):
s = (vcos(α) + w)t1 (3)
We can calculate t1 from equation (2). It has two solutions:
t1 = 0, and t1 = 2vsin(α)/g
The solution t1=0 means that at time 0 the two cats have the same y-coordinates,
which is true, but not useful for us.
The second solution is useful, and we can substitute it into (3):
s = (vcos(α) + w)2vsin(α)/g
Substituting actual numbers
s = (1.4×cos(13°) + 2)×2×1.4×sin(13°)/9.81 = 0.216
So Chloe should jump when Luna is about 21.6 cm away.
Please note that this answer makes no sense in reality, as 21 cm is about the length of a cat.
To solve this problem we had to treat both cats as if they were mere dots.