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Physics 13 Online
OpenStudy (anonymous):

A parachutist with a camera, both descend- ing at a speed of 10 m/s, releases the camera at an altitude of 28.4 m. What is the magitude of the velocity of the camera just before it hits the ground? The acceleration of gravity is 9.8 m/s2 and air friction is negligible. Answer in units of m/s

OpenStudy (aaronq):

1. Draw a picture |dw:1413064988727:dw| 2. write equation (kinematic equations apply here because body is under constant acceleration) \(\large \sf v_f^2=v_i^2+2a\Delta y\) \(\large \sf v_f^2=(-10~m/s)^2+2(-9.8~m/s^2)(28.4~m)\) 3. Solve equation

OpenStudy (aaronq):

28.4 m should have been negative, -28.4 m

OpenStudy (anonymous):

@aaronq How do I find how long the camera will take to reach the ground?

OpenStudy (aaronq):

use another one of the kinematic equations: \(\large \sf v_f=v_i+a\Delta t\)

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