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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?

Respuesta :

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

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