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

[Tikiwreck]

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?

 

[the_d]

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

 

[kyle8921]

A. To calculate the deceleration of the coyote, we can use the formula: d = (v0 + vf)t/2, where d is the distance (500m), v0 is the initial velocity (0m/s), vf is the final velocity (unknown), and t is the time (6 seconds). Rearranging the formula, we get vf = 2d/t - v0. Plugging in the values, we get vf = 166.67 m/s. Therefore, the deceleration of the coyote is 166.67 m/s2.

B. To find the maximum height reached by the coyote on his trip back up, we can use the formula: h = v0t + (1/2)at2, where h is the maximum height (unknown), v0 is the initial velocity (166.67 m/s), a is the acceleration due to gravity (-9.8 m/s2), and t is the time (5 seconds). Rearranging the formula, we get h = v0t - (1/2)at2. Plugging in the values, we get h = 416.675 meters. Therefore, the maximum height reached by the coyote is 416.675 meters.

C. To find the final velocity of the coyote as he reaches the ground for the second time, we can use the formula: vf = v0 + at, where vf is the final velocity (unknown), v0 is the initial velocity (166.67 m/s), a is the acceleration due to gravity (-9.8 m/s2), and t is the time (11 seconds). Plugging in the values, we get vf = 0 m/s. Therefore, the final velocity of the coyote as he reaches the ground for the second time is 0 m/s.