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

Answer:

$\alpha =54.7$º

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 $y_{o}=0$

$0=v_{oy}^2-2(9.8m/s^2)(3.4m)$

$v_{oy}=\sqrt{2(9.8m/s^2)(3.4m)}=8.16m/s$

Knowing that

$v_{oy}=v_{o}sin\alpha$

$sin\alpha =\frac{v_{oy} }{v_{o} }$

$\alpha =sin^{-1}(\frac{v_{oy}}{v_{o}})=sin^{-1}(\frac{8.16m/s}{10m/s})=54.7$º