Some enterprising physics students working on a catapult decide to have a water balloon fight in the school hallway. The ceiling is 3 m in height, and the balloons are launched at a velocity of 10 m/s. the acceleration of gravity is 9.8 m/s/s.
At what angle must they be launched to just graze the ceiling? Answer in units of degrees.
2007-11-26 12:15:31 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
use energy in the y
.5*m*vy^2=m*g*h
vy=sin(th)*10
sin(th)=sqrt(2*9.8*3/100)
th=50 degrees
j
2007-11-26 12:18:54 · answer #1 · answered by odu83 7 · 0⤊ 0⤋
For a projectile projected with velocity U at angle O with the horizontal,
Maximum height attained=Hmax =U^2 sin^2O/2g
sinO = sq rt [2gHmax / U^2 ]
sinO = sq rt [2*9.8*3 / 10*10 ] = sq rt 0.588=0.7668
angle of projection = O = 50.06 degree
The balloons be launched at 50.06 degree with the horizontal, to just graze the ceiling
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2007-11-26 20:32:56 · answer #2 · answered by ukmudgal 6 · 0⤊ 0⤋