Will you please be able to explain the solution to the problem in 28 steps. The

Question

Will you please be able to explain the solution to the problem in 28 steps.

The determined coyote is out more in pursuit of the elusive a pair of Acme jet powered roller skates, which provide a constant horizontal acceleration of 15 m/s^2. The coyote starts at rest 70.0 m from the brink of a cliff at the instant the roadrunner zips past him in the direction of the cliff.

a.) If the roadrunner moves with constant speed, determin the minimum speed he must have so as to reach the cliff before the coyote.

At the edge of the cliff, the roadrunner escapes by making a sudden turn, while the coyote continues straight ahead. His skates remain horizontal and continue to operate while his is in flight so that his acceleration while in the air remains constant

b) The cliff is 100 m above the flat floor of a wide canyon. determine where the coyote lands in the canyon.

c) Determine the component's of the coyote's impact velocity

Explanation / Answer

A.) For this part, you must calculate the amount of time it will take the coyote to reach the edge of the cliff using the formula x=1/2 at^2.  If you rearrange this for time you get t=sqrt(2x/a), x=lateral distance being 70m, a = acceleration being 15m/s^2.  This gave me t=3.055s.  This value can now be used to calculate an average velocity for the roadrunner by using v=x/t, or v=70/3.055.  V=23m/s

B.)  You will need to calculate the amount of time the coyote will fall using the same t=sqrt(2x/a) formula again, but this time a=9.8m/s^2 and x=the height of the cliff.  Plug this value into the formula x=1/2at^2+Vot, where x will be the lateral canyon distance, a=15m/s^2, Vo=the velocity of the coyote at the point at which he goes off the cliff (you will need to calculate that by multipling the time from part A by the acceleration (3.055x15)) and t=the time you have just calculated.  I got 360m.

C.)For this part you will need to calculate the final velocities for each component, x and y, using the formula Vf=Vo+at, where Vf is the final velocity, Vo in the initial velocity (only applicable in the x direction), a=the acceleration in the direction of the component (either 9.8 for y or 15 for x) and t=the time that was calculated in Part B.  I got 113.6m/s for the final velocity in the x direction and 44.3m/s for the final velocity in the y-direction.

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