Wile E. Coyote is holding a “HEAVY DUTY ACME ANVIL” on a cliff that is 40.0 meters high. The Roadrunner (beep-beep), who is 1.0 meter tall, is running on a road toward the cliff at a constant velocity of 10.0 m/s. Wile E. Coyote wants to drop the anvil on the Roadrunner’s head. How far away should the Roadrunner be when Wile E. drops the anvil?
I got a time of 2.89, using the position in the y direction .. then multiplied by the 10 m/s velocity (x=vt) to get 28.9.. but the choices are 28.2 and 28.6??
y = 39m - 1.0 m (Remember the Roadrunner's height!)= 38m
g = 9.8 m/s/s
vo = 0 m/s (dropped from rest)
y = 0.5gt^2
t = sqrt(38/4.9) = approx. 2.78s (rounded from 2.784798384)
So now you have t.
x = vt, and v = 10 m/s
x = 10(2.78) gets you 27.8 meters.
My homework says the answer should be 28.6, though.
Actually, this is how I got it: (based on my worksheet)
dy=40.0m
= (40.0m/4.9)
= squareroot(8.16) = 2.857 round off = 2.86 (10m)
= 28.6 m
Boom! 28.6m is the answer.
y = 39m
g = 9.8 m/s/s
vo = 0 m/s (dropped from rest)
y = 0.5gt^2
t = sqrt(39/4.9) = approx. 2.82s (rounded from 2.821202523)
So now you have t.
x = vt, and v = 10 m/s
x = 10(2.8) gets you 28.2 meters.
My homework says the answer should be 28.6, though.
Well, well, well, looks like Wile E. Coyote is up to his old shenanigans again! Trying to drop an anvil on the Roadrunner, huh? Classic.
Now, let's crunch some numbers. We know that the anvil will fall for a certain amount of time, and during that time, the Roadrunner will be running towards the cliff.
First, let's find out how long it takes for the anvil to fall. The height of the cliff is 40.0 meters, and if we assume no air resistance (because, you know, physics in cartoons), we can use the equation:
h = 1/2 * g * t^2
where "h" is the height, "g" is the acceleration due to gravity, and "t" is time. Plugging in the values, we get:
40.0 = 1/2 * 9.8 * t^2
Now, solving for "t," we find that it takes approximately 2.02 seconds for the anvil to hit the ground.
Next, since the Roadrunner is running towards the cliff at a velocity of 10.0 m/s, we can find how far he travels in 2.02 seconds using the formula:
x = v * t
where "x" is the distance, "v" is the velocity, and "t" is time. Plugging in the values, we get:
x = 10.0 * 2.02
So, the Roadrunner will be approximately 20.2 meters away from the cliff when the anvil is dropped.
So, according to my calculations, the answer should be around 20.2 meters, which is not given in the choices you mentioned. It seems like we have a mystery on our hands!
To solve this problem, we can use the equations of motion in the y-direction.
First, let's find the time it takes for the anvil to fall from the cliff. We can use the equation:
y = y0 + v0*t + (1/2)*a*t^2,
where
y = final position (0, since the anvil is at ground level),
y0 = initial position (40.0 m),
v0 = initial velocity (0, since the anvil is initially at rest),
a = acceleration due to gravity (-9.8 m/s^2),
t = time.
Substituting in the values, we get:
0 = 40.0 - (1/2)*9.8*t^2.
Simplifying this equation, we have:
19.6*t^2 = 40.0.
Dividing both sides by 19.6, we get:
t^2 = 2.04.
Taking the square root of both sides, we find:
t = √(2.04) ≈ 1.43 s.
Now, since the Roadrunner is 1.0 meter tall and moving at a constant velocity of 10.0 m/s, we can find the distance it travels during this time interval. The equation to use is:
x = v*t,
where
x = distance,
v = velocity in the x-direction (10.0 m/s),
t = time (1.43 s).
Substituting the values, we find:
x = 10.0 m/s * 1.43 s ≈ 14.3 m.
So the Roadrunner should be approximately 14.3 meters away from Wile E. Coyote when he drops the anvil.
The discrepancies between your answer and the given choices could be due to rounding errors or different approaches to solving the problem.
The anvil falls 39m
39=1/2 g t^2
t=sqrt(78/9.8)
I don't get your time.