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December 22, 2014

December 22, 2014

Posted by **Kayla** on Tuesday, January 22, 2013 at 12:04am.

- Physics -
**Devron**, Wednesday, January 23, 2013 at 3:13pmVi=0 in the Y-direction of the bag going down, a=g=9.8m/s^2, d=40m. Even if it is rising at 5m/s, the velocity going down will not no be affected by the velocity going up. Since the velocity is 0, initially, and it accelerates down at a rate of 9.8m/s^2, multiply 9.8m/s^2 by 0.250s to give you the velocity at this time interval; notice that one of the s's cancel to give you units of m/s which is velocity at that time. In one second, the sand bag accelerated by 9.8m/s^2. Since the initial velocity (Vi) is 0 at the time of release, in one second the velocity is equal to gravity=9.8m/s. Use the formula (Vf)^2=(Vi)^2 +2ad to find the final velocity (Vf). Since Vi=0, the formula turns into Vf=(2ad)^1/2=(2(9.8m/s^2)(40m))^1/2. Once you compute the Vf, plug in Vf, Vi=0, and a=9.8m/s^2 into one of your other kinematic equations (Vf=Vi+at) and solve for time (t). Since the bag is released at 40m, it only reaches a max height of 40m.

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