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

A hotair balloonist, rising vertically with a constant speed of 5.00 m/s, releases a sadndbag at the instant the balloon is 40.0 m above the ground. after it is released, the sandbag encounters no appreciable air drag. a. compute the position and velocity of the sandbag at .250 s and 1.00 s after its release. b. how many seconds after its release will the bag strike the ground? C. how fast is it moving as it strikes the ground? d. what is the greatest height above the ground that the sandbag reaches?

- Physics -
**Devron**, Wednesday, January 23, 2013 at 3:13pm
Vi=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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