Question

A water balloon launcher launches a balloon straight up at a flying kite. If the kite is hovering 75.0 meters off the ground, and the balloon hits it on its upward trajectory with a velocity of 12.2m/s, a) What was the launch velocity of the balloon

Answer 1

Launch velocity of the balloon is 40.23 m/s.

Step-by-step explanation:

It is given that,

Height above ground, h = 75 m

Velocity of balloon on its upward trajectory, u = 12.2 m/s

We need to find the launch velocity of the balloon. Let it is equal to v. Applying the conservation of energy as :

`mhg+(1)/(2)mu^2=(1)/(2)mv^2`

`hg+(1)/(2)u^2=(1)/(2)v^2`

`75(9.8)+(1)/(2)(12.2)^2=(1)/(2)v^2`

`v=40.23\ m/s`

So, the launch velocity of the balloon is 40.23 m/s. Hence, this is the required solution.