posted by Anonymous .
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.
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.
How did you get the pellet equasion? And how did you simplify it without knowing the value of g?
He used g = 9.8
(1/2) g = 4.9
For constant acceleration
x = Xo + Vo t + (1/2) a t^2
height = initial height + 30 t - 4.9 t^2
O right, gravity is 9.8
So, because of the constant speed of the ballon, we considere (1/2)*g*t² for the balloon to be 0 ?