Question

Determined to test gravity, a student walks off the Burj Khalifa (aka really tall building) in Dubai, which is

829 m high, and falls freely. His initial velocity is zero. Jetpack Woman leaves the roof 5 s later to save
the student with an initial velocity downward and then is in free-fall. In order both to catch the student
and to prevent injury to her, Jetpack Woman should catch the student at a sufficiently 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 (4 G). (You may assume that Jetpack Woman has superpowers, which allow her to change
velocity instantaneously.) How high above the ground must Jetpack Woman catch the student?

Answer 1

y = 165.8 m

Step-by-step explanation:

This is an exercise in kinematics that we must solve in parts,

Let's start with the final part where we have more data, the final velocity of the student and jet is zero, the upward acceleration is

a = 4g = 4 9.8 = 39.2 m /s²

Let's find the initial velocity and the height

v² = v₂² + 2 a y

0 = v₂² + 2 ay

2ay = - v₂²

v₂² = 2a y

Let's write the equation for the first part

student

v₂² = v₀² - 2 g (y₀ -y)

in this case the student jumps v₀ = 0

v₂² = 2 g (y₀ -y)

the total height is y₀ = 829 m

let's solve the system of equations by equating

2ay = 2 g (y₀ -y)

2a y = 2 g y₀ - 2g y

2 y ( g + a) = 2 g y₀

y = yo g/(g +a)

a= 4 g

y = yo 1/5

calculate

y = 829 /5

y = 165.8 m