A relief package is released from a helicopter at 1600 feet

A relief package is released from a helicopter at 1600 feet. The height of the package can be modeled by the equation: h(t)   -16t squared + 1600 where h is the height of the package in feet and t is the time in seconds. The pilot wants to know how long it will take for the package to hit the ground if the helicopter copilot THREW the package down at an initial velocity of 80 ft/sec

Solution

Had the package been just released, it would have hit the ground in 10 sec ( when -16t² + 1600 = 0,   t² = 100 and t = 10) . However, since the package has been thrown with an initial velocity of 80 ft/sec , its motion is governed by the equation d = vt + (1/2) at² where d is the distance, t is time, v is initial velocity and a is acceleration. Here, a = g = Its value is 9.8 m/s² or. 32.152ft/s² . Therefore, we have 1600 = 80t + 16.076 t²    or, 16076t² + 80000t - 1600000 = 0. Therefore t = [ - 80000±{ ( 80000)² - 4(16076)(- 1600000)} ] / 2* 16076 = [ - 80000 ± ( 6400000000 + 102886400000)] /32152 = ( -80000 ± 330585)/ 32152 ( approx.) = 250585/32152 = 7.79 seconds approximately ( as t cannot be negative)