A base jumper dives off a mountain top. flying with her wing suit through a cany

Question

A base jumper dives off a mountain top. flying with her wing suit through a canyon. She wants to fly through a narrow crack between beaks of rock. 600 m below the starting point, and at a an angle of alpha = 65 degree below the horizon. She can fly at 160 km/h in still air. To day, the sunny a thermal lift of 5 m/s upwards. Make a sketch containing all the quantities relavent to the problem. Include units. What is the horizontal width w of the planned flight from the mountain top to the narrow crack? What is the angle between heading and bearinn of the base jumper? flow fast docs the base jumper fly relative to the mountain, taking the thermal drift into account? How long dos the flight take?

Explanation / Answer

heading is the direction in which the jumper is pointing

in this case it would be 65 degree below horizontal

bearing will be the direction in which she is actually moving

it can be found by taking thermal lift into account

160 km/hr=44.44 m/s
at any time t before reaching ground,

her vertical speed=initial vertical speed+acceleration due to gravity*t

=44.44*sin(65)+9.8*t

=40.28+9.8*t

due to thermal lift, her vertical speed will become 40.28+9.8*t-5=35.28+9.8*t

her horizontal speed will remain constant at 44.44*cos(65)=18.78 m/s

so at any time point, her bearing with vertical will be arctan(horizontal speed/vertical speed)

=arctan(18.78/(35.28+9.8*t))

let time taken for her flight is T seconds

then 600=44.44*sin(65)*T-5*T+0.5*9.8*T^2

==>600=35.28*T+4.9*T^2

solving for T, we get T=8.0635 seconds

then her bearing with vertical at the end of the flight=arctan(18.78/(35.28+9.8*8.0635))=9.33 degrees

hence her bearing with horizonal direction=90-9.33=80.67 degrees

then angle between bearing and heading=80.67-65=15.67 degrees

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