Answer:
t = 1.42 s and d = 35.5 m
Step-by-step 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+(1)/(2)at^2\\\\\text{Here, u = 0 and a = g}\\\\d=(1)/(2)gt^2\\\\t=\sqrt{(2d)/(g)} \\\\t=\sqrt{(2* 10)/(9.8)} \\\\t=1.42\ s](https://img.qammunity.org/2021/formulas/physics/high-school/pekak93x2780b18ovc2l5gvq9koo0s12m5.png)
Let d is the distance traveled. So,
d = vt
d = 25 m/s × 1.42 s
d = 35.5 m