What angle tilt the ball launc
Please help me. I don't know that how to do it?
What happened?The picture disappear very suddenly.
- ?Lv 71 decade agoFavorite Answer
Let the launch angle be a. The vertical and horizontal portion of the speed would be 3sin(a) and 3cos(a), respectively. Let the time for the ball to reach the hole be t. So t=2.5/3cos(a) . The acceleration on the tilt flat board is gsin(20o). Since the ball goes up and then back to the bottom, the vertical displacement is zero, i.e. 0=3sin(a)t-gsin(20o)t2/2, or 3sin(a)=gsin(20o)t/2. Since g=9.8 m/s2, the equation can be rewritten by replacing t with eq. .
3sin(a)=4.9sin(20o)*2.5/3cos(a), 2sin(a)cos(a)=0.931, sin(2a)=0.931, 2a=Asin(0.931)=68.6o, a=34.3o