Calculating Coyote's Fall and Flight: Motion Question & Final Velocity

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SUMMARY

The discussion revolves around a physics problem involving a coyote falling from a height of 500 meters and utilizing a rocket-powered backpack. After 6 seconds of free fall, the coyote experiences constant deceleration to achieve a gentle landing with zero velocity. The deceleration can be calculated using the formula a = Vo/t^2. Following the landing, the coyote is propelled back into the air for 5 seconds before the rocket runs out of fuel, allowing for the calculation of the maximum height reached during the ascent and the final velocity upon descending again.

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Confusing Motion Question..Please Help!

Question:
The coyote in his relentless attempt to catch the elusive roadrunner loses his footing and falls from a sharp cliff, 500meters above ground level. After 6seconds of free fall the coyote remembers he is wearing his ACME rocket-powered backpack, which he turns on.

A. The coyote comes to the ground with a gentle landing (i.e. zero velocity). Assuming constant deceleration, find the deceleration of the coyote.

B. Unfortunately for the coyote, he is unable to shut down the rocket as he reaches the ground. Consequently, he is propelled back up into the air. After 5seconds the rocket runs out of fuel. Find the maximum height reached by the coyote on his trip back up.

C. What is his final velocity as he reaches the ground for the second time?
 
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You are given the distance the coyote falls and the time of the fall and the final velocity, with this information you can find the initial velocity of the coyotes fall Vo = d/t, and then you are able to find the acceleration or deceleration of the coyote using your answer from the previous eq'n in the formula for acceleration, a = Vo/t^2 for part a
 

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