Denote the elevation angle as `alpha(t)` (in radians, `t` is in hours). Denote the known speed as `V` and the (unknown) horizontal distance between the airplane and the kangaroo that corresponds to `t=0` as `L.` Then the horizontal distance at any time `t` is equal to `L - Vt.`
This way the tangent of `alpha(t)` is `2/(L - Vt),` so `alpha(t) = arctan(2/(L - Vt)).`
The increase of the angle of elevation is the derivative of `alpha(t),` and we are interested of its value for `t` such that the height `2mi,` the horizontal distance `L - Vt` and the distance `3mi` form a right triangle. In other words, `L - Vt = sqrt(3^2-2^2) = sqrt(5) (mi).`
Find the derivative:
`d/dt(alpha(t)) = 1/(1+(2/(L - Vt))^2)*(2V)/((L - Vt)^2) =(2V)/((L - Vt)^2+4).`
When `L - Vt = sqrt(5)` it will be `(2V)/9 approx 133` (radians per hour). In radians per minute it will be `60` times less, about 2.2. This is the answer.