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
here
height = initial height + 30 t - 4.9 t^2

O right, gravity is 9.8

thanks!

So, because of the constant speed of the ballon, we considere (1/2)*g*t² for the balloon to be 0 ?

To determine the height at which the pellet and balloon meet, you can set up the kinematic equations for both the pellet and the balloon and then solve for the common height.

Let's define the variables first:
- h_balloon: height of the balloon above the ground
- h_pellet: height of the pellet above the ground
- t: time when the pellet and balloon have the same altitude

For the balloon:
The balloon is ascending straight up at a constant speed of 7.0 m/s. Since the initial height of the balloon is 12.0 m above the ground, we can represent the height of the balloon as:
h_balloon = 12.0 + 7.0t -- Equation 1

For the pellet:
The pellet is projected straight up from ground level with an initial speed of 30.0 m/s. The height of the pellet can be represented as:
h_pellet = 30.0t - (1/2)gt^2 -- Equation 2

In Equation 2, 'g' represents the acceleration due to gravity, which is approximately 9.8 m/s^2.

To find the height at which the pellet and balloon meet, we need to find the value of 't' when h_balloon = h_pellet. So, we can set Equation 1 equal to Equation 2:

12.0 + 7.0t = 30.0t - (1/2)gt^2

Now, we can solve this equation to find the value of 't' and then substitute it back into Equation 1 or Equation 2 to find the corresponding height.

After simplifying the equation, it becomes a quadratic equation:
(1/2)gt^2 - 23t + 12 = 0

You can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)

where a = (1/2)g, b = -23, and c = 12.

Once you find the values of 't', substitute them back into Equation 1 or Equation 2 to find the corresponding heights above ground level where the pellet and the balloon meet.