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waldorfchick1 · Sep 21, 2008, 07:59 PM

I really need help with this question..

a hot air balloon is ascending straight up at a consant speed of 7.0 m/s. When the baloon is 12.0 m above the ground, a gun fires a pellet straight up from the ground level with an initial speed of 30.0 m/s. along the paths of the balloon and the pellet, there are two places where each of them has the same altitude at the same time. How far above ground level are these places?

these are the formulas if uve forgotton:

V=Vo+at
x=1/2 (Vo+V)t
x= Vot +1/2at^2
V^2=Vo^2 +2ax

ebaines · Sep 22, 2008, 05:53 AM

The pellet will pass the balloon on its way up, and again as it falls back to earth. Set up the equations for the balloon and pellet in terms of time t, solve for t (you'll get two solutions), and then plug the result back into one or the other equation of motion to find the altitude of either the balloon or the pellet at that point in time. I'll get you started: the equation for the balloon comes from x = Vot + 1/2at^2, but in this case a = 0 (since the balloon moves at constant velocity), and the initial displacement is 12 m, so you get: x = 12m + 3(m/s)*t. The pellet's motion from the same equation is: x = 30 m/s *t - 1/2*g*t^2. Set these equal, solve for t, and then shove that value for t back into one or the other of the equations of motion.