Part 2: The actual execution At a distance of L-200m, from the coyote\'s initial

Question

Part 2: The actual execution At a distance of L-200m, from the coyote's initial position, there is an abrupt ending to the road, on which there is a cliff. The roadrunner notices the end while being 2m away and makes a complete stop, ending right at the edge of the cliff d) What is the average acceleration of the roadrunner during the stop? (Magnitude and direction) In contrast, the coyote cannot control the rocket, so being unable to stop he keeps moving and falls from the cliff (of height H 100 m), while his propulsion is still on,and the rocket stays in a horizontal orientation e) What is the total velocity of the coyote right at the edge of the cliff, before he begins the fall? f) How long will it take the coyote to hit the bottom of the cliff? E) What will his total velocity be the moment he hits the bottom? (Give answer in Cartesian and polar coordinates) What is his speed right before the crash? h) How far from the edge of the cliff will the coyote crash? i) For the journey between the edge of the cliff and the crash site, and taking the edge as the origin of coordinates, calculate the coyote's a. total displacement b. average velocity c average acceleration Note: Assume the coyote comes to a complete stop upon hitting the ground at the end of the journey Caution: you are dealing with 2D vector quantities here. Give all vectorial answers in Cartesian and polar coordinates.

Explanation / Answer

d) Here initial velocity u = 36 km/h = 36*5/18 m/s= 10 m/s, final velocity v = 0

now, using third equation of motion,

a = u^2/2s = 10^2/(2*2) = 25 m/s^2

e) data from partA may be required for acceleration of coyote. v = sqrt(2as) where s = 200+2 = 202 m

f) by second equation of motion, h = 0.5 gt^2

t = sqrt(2*100/9.8) = 4.52 s

g) This part also needs data from last part. However, following concept will be applied.

Let the answer in part e be b m/s

vertical velocity = -gt = -9.8*4.52 = -44.3 m/s

v = b i -44.3 j m/s

speed = sqrt(b^2+44.3^2)

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