A small airplane flying horizontally with a speed of 180 mi/hr at an altitude of
ID: 1493358 • Letter: A
Question
A small airplane flying horizontally with a speed of 180 mi/hr at an altitude of 400 ft above a remote valley drops an emergency medical package at A. The package has a parachute which deploys at B and allows the package to descend vertically at the constant rate of 6 ft/s. If the drop is designed so that the package is to reach the ground 37 seconds after release at A, determine the horizontal lead L so that the package hits the target. Neglect atmospheric resistance from A to B 180 mi/hr 400 TargetExplanation / Answer
180mph * 22fps/15mph = 264 ft/s
For the plane to travel L takes time t = L / 264fps = 0.0038L
During that time, the package falls d = ½at² = ½ * 32ft/s² * (L/264fps)²
or d = 0.00023*L² where d and L are in feet.
It falls the remaining distance h = 400ft - d in time
T = h / 6ft/s = (400 - d) / 6 = (400 - 0.00023L²) / 6 = 66.7 - 0.000038L²
We are given that t + T = 37 s, so
0.0038L + 66.7 - 0.000038L² = 37
0.000038L² - 0.0038L - 29.7 = 0
quadratic in L, solutions at L = -832 ft, 931 ft
L = 931 ft.........Ans.