I need help in how to solve this including the formula used. Determined to test

Question

I need help in how to solve this including the formula used. Determined to test gravity a student walks off the BURJ Khalifa in Dubai, which is 829m high and falls freely. His initial velocity is zero. JETPack woman leaves the roof with an initial velocity downward and then is in freefall. In order to both catch the student and to prevent injury to her, the jetpack woman should catch the student at a sifficiently great height and arrive at the ground with zero velocity. The upward acceleration that accomplishes this is provided by Jetpack woman's jet pack, which she turns on just as she catches the student, before then the student is in free fall. to prevent discomfort to the student the magnitude of the acceleration is limited to four times gravity 4g. HOW HIGH ABOVE THE GROund must jetpack woman catch the student?
Also, what would the velocity vs. time graph, and acceleration vs. time graph look like?

Explanation / Answer

h1 = the distance traveled befor catching:

h1 = v^2/(2*g)

h2 = the distance traveld after catching:

h2 = v^2/(2*4g)

h1+h2 = 829

v^2/(2g) + v^2/(2*4g) = 829

v^2 * (1/(2g) + 1/(8g)) = 829

v^2 * (5/(8g)) = 829

v^2 = 829*8*9.81/5

v^2 = 13011.984

h2 = v^2/(2*4g) = 13011.984/(8*9.81) = 165.8 m

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