
Andy C. answered 04/29/18
Tutor
4.9
(27)
Math/Physics Tutor
Y is the distance from the balloon from the ground.
By pythagorean theorem , the distance from the rangefinder
to the balloon is
sqrt( y^2 + 250000) = F(y)
dF/dt = 1/2( y^2 + 250000) ^(-1/2) ( 2y) (dy/dt)
F'(600) =
1/2( 600^2 + 250000) ^(-1/2) ( 2*600) ( 140)
= (1/2)(1200)(140) / sqrt( 600^2 + 250000)
= 84000 / sqrt( 360000 + 250000)
= 84000 / sqrt(610000)
107.551 ft/sec
By pythagorean theorem , the distance from the rangefinder
to the balloon is
sqrt( y^2 + 250000) = F(y)
dF/dt = 1/2( y^2 + 250000) ^(-1/2) ( 2y) (dy/dt)
F'(600) =
1/2( 600^2 + 250000) ^(-1/2) ( 2*600) ( 140)
= (1/2)(1200)(140) / sqrt( 600^2 + 250000)
= 84000 / sqrt( 360000 + 250000)
= 84000 / sqrt(610000)
107.551 ft/sec