Question

Some enterprising physics students working on a catapult decide to have a water balloon fight in the school hallway. The ceiling is of height 3.4 m, and the balloons are launched at a velocity of 10 m/s. The acceleration of gravity is 9.8 m/s 2 . At what angle must they be launched to just graze the ceiling? Answer in units of ◦

Answer 1

`\alpha =54.7`º

Step-by-step explanation:

From the exercise we have our initial information

`y=3.4m\\v_(o)=10m/s\\g=-9.8m/s^2`

When the balloon gets to the ceiling its velocity at that moment is 0 m/s. Being said that we can calculate velocity at the vertical direction

`v_(y)^2=v_(oy)^2+ag(y-y_(o))`

Since
`v_(y)=0` and
`y_(o)=0`

`0=v_(oy)^2-2(9.8m/s^2)(3.4m)`

`v_(oy)=√(2(9.8m/s^2)(3.4m))=8.16m/s`

Knowing that

`v_(oy)=v_(o)sin\alpha`

`sin\alpha =(v_(oy) )/(v_(o) )`

`\alpha =sin^(-1)((v_(oy))/(v_(o)))=sin^(-1)((8.16m/s)/(10m/s))=54.7`º