Posted by Anonymous on .
A hot air balloon is ascending straight up at a constnat speed of 7.0 m/s. When th balloon is 12.0 m above the ground, a gun fires a pellet straight up from ground elvel 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?
I'm thinking I should be setting two kinematics equations equal to each other, but I don't know which ones.

Physics 
drwls,
For the balloon,
Y = 7 t + 12
where t is the time the pellet is fired.
For the pellet,
Y = 30 t  (g/2) t^2
= 30 t  4.9 t^2
Solve for the two t's when the Y's are equal.
4.9 t^2  23 t +12 = 0
Note that this is a quadratic equation with two roots.
Then use either Y equation to get the corresponding altitudes at those two times. 
Physics 
Anonymous,
How did you get the pellet equasion? And how did you simplify it without knowing the value of g?

Physics 
Damon,
He used g = 9.8
(1/2) g = 4.9
For constant acceleration
x = Xo + Vo t + (1/2) a t^2
here
height = initial height + 30 t  4.9 t^2 
Physics 
Anonymous,
O right, gravity is 9.8
thanks! 
Physics 
Kayline,
So, because of the constant speed of the ballon, we considere (1/2)*g*tÂ² for the balloon to be 0 ?