A horse 1 m tall running towards a tree at a constant velocity of 20m/s, there was a baboon on the tree which is 3m tall, if the baboon falls exactly on
the back of the horse and sped away with baboon at its back.
Calculate
I) how far was the horse from the tree and the time the baboon
falls on the horse.

The horse's position on the ground at time t is

x = (20 m/s) t

The baboon's height from the ground at time t is

y = 3 m - 1/2 g t²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The baboon falls and lands on the horse, so that the two animals meet when the baboon's height is 2 m from the ground, which happens after

2 m = 3 m - 1/2 g t²

1/2 g t² = 1 m

t² = (2 m) / (9.80 m/s²)

t ≈ 0.452 s

In this time, the horse reaches the tree, so its distance from it is

(20 m/s) * (0.452 s) ≈ 9.04 m

with 5 seconds, no, but over time if the police car stays at 65m/s ad the speeding car stays at 60, then obviously yes, but the answer to your question would be, no.

hope this : )