a hot air ballon is 2.5m high, moving at 2/ms. when someone throws a camera at 13m/s to the baloon. at which height do they meet?

To find the height at which the camera and the hot air balloon meet, we need to determine the time it takes for the camera to reach the balloon. The time can be found by dividing the horizontal distance covered by the camera by its horizontal velocity.

Let's denote the initial height of the balloon as H_balloon = 2.5 meters.
The horizontal velocity of the hot air balloon (and the camera) is 2 m/s.
The horizontal distance covered by the camera is given as 13 meters.

Using the formula: distance = velocity × time, we can rearrange it to find time.

time = distance / velocity

The time taken by the camera to cover the horizontal distance of 13 meters is:

time = 13 meters / 2 m/s = 6.5 seconds

Now that we have the time it takes for the camera to reach the balloon, we can calculate the height at which they meet.

The vertical motion of the camera can be described by the equation S_camera = Ut + (1/2)gt^2, where:
S_camera is the vertical displacement of the camera (height)
U is the initial vertical velocity of the camera
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time taken (6.5 seconds)

Using this equation and considering the initial height of the balloon as 2.5 meters, we can solve for the height at which they meet.

S_camera = Ut + (1/2)gt^2
S_camera = 0 + (1/2) * 9.8 * (6.5)^2

S_camera = 0 + 3.9 * 42.25
S_camera = 0 + 164.775

S_camera = 164.775 meters

Therefore, the camera and the hot air balloon meet at a height of approximately 164.775 meters above the initial height of the balloon.