Question

You are wearing your rocket pack (m total = 75 kg) that accelerates you upward from rest at a constant 8m/s^2 . While preparing to take pictures of the view, you drop your camera 3.0 seconds after liftoff. Two seconds after you drop the camera,

(a) what is the camera’s velocity,
(b) how far are you from the camera,
(c) create a velocity-time graph for the camera on a sheet of paper.

Answer 1

below

Step-by-step explanation:

The camera's initial upward velocity will be the same as yours at that instant

Vf = at = 8 * 3 = 24 m/s upward

THEN (after you drop it) , its velocity is given by:

Vf = Vo + at where a = - 9.81 m/s^2

Vf = 24 + (-9.81)(2) = 4.38 m/s upward = ~ 4.4 m/s

b) in two seconds YOU travel (from the point you dropped your camera)

d = vo t + 1/2 a t^2

= 24 (2) + 1/2 ( 8)(2^2) = 64 meters

and your camera travels d = Vot + 1/2 at^2

24 (2) + 1/2(-9.81)(2^2) = 28.38 m

a difference of 64 - 28.38 = 35.62 m