Posted by **Anonymous** on Friday, January 2, 2009 at 10:44pm.

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**, Friday, January 2, 2009 at 11:15pm
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**, Friday, January 2, 2009 at 11:39pm
How did you get the pellet equasion? And how did you simplify it without knowing the value of g?

- Physics -
**Damon**, Saturday, January 3, 2009 at 5:16am
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**, Saturday, January 3, 2009 at 2:17pm
O right, gravity is 9.8

thanks!

- Physics -
**Kayline**, Wednesday, October 13, 2010 at 9:42am
So, because of the constant speed of the ballon, we considere (1/2)*g*tē for the balloon to be 0 ?

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