A package is dropped from a helicopter moving upward at 1.5 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released if air resistance is negligible?

Well we know acceleration from free fall due to gravity is 9.8m/s^2

Lay out

S = displacement is what we need

U

V = 1.5m/s

A = 9.8m/s2

T = 16.0s

Use the equation s=vt-1/2at^2

Where a = acceleration t= time and v= velocity

Sub in the values to get displacement or height from ground

= -1230.4 metres which would be positive as you’re measuring distance (scalar quantity) so it’s 1230.4 metres

Ithink the answers is b i hope it my friend message me if i’m wrong and i’ll change my answer and fix it for you

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2