A horse 1 m tall running towards a tree at a constant velocity of 20m/s,

Physics

1 answer:

kykrilka [37]3 years ago

7 0

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

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