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V125BC [204]
2 years ago
5

A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the

roadrunner using a net dropped from an overpass that is 10m high, how much time before the roadrunner is under the overpass, should the coyote drop the net? How far away from the overpass is the roadrunner when the coyote drops the net?
Physics
1 answer:
andreyandreev [35.5K]2 years ago
5 0

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and  how far away from the overpass is the roadrunner when the coyote drops the net.

d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

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