Another scheme to catch the roadrunner has failed. A safe falls from rest the top of a 30.2 m high cliff toward Wiley Coyote, who is standing at the base. Wiley first notices the safe after it has fallen 17.4 m.
How long does he have to get out of the way? The acceleration of gravity is 9.8 m/s^. Answer I units of s.
35
To determine how long Wiley Coyote has to get out of the way, we can use the equations of motion and the information given in the problem.
The given information is:
Initial position (s0) = 0 m (since the safe falls from rest)
Final position (s) = 17.4 m (the height at which Wiley first notices the safe)
Acceleration due to gravity (a) = 9.8 m/s^2
Distance fallen (d) = 30.2 m (height of the cliff)
We can start by using the equation of motion: s = s0 + v0t + 0.5at^2.
In this case, s0 = 0 (since the safe falls from rest), and v0 = 0 (since it starts from rest).
Substituting the given values into the equation, we have:
17.4 = 0 + 0 + 0.5 * 9.8 * t^2.
Simplifying the equation, we have:
17.4 = 4.9t^2.
Dividing both sides of the equation by 4.9, we get:
t^2 = 3.55.
To find the time (t), we can take the square root of both sides of the equation:
t = √(3.55).
Using a calculator, we find that √(3.55) ≈ 1.888.
Therefore, Wiley Coyote has approximately 1.888 seconds to get out of the way before the safe reaches him.
2.01s
0.5g*t^2 = 30.2-17.4 = 12.8 m. Above gnd
4.9*t^2 = 12.8.
t^2 = 2.61 s.