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I need physics help please?
A soccer ball is kicked with a speed of 15.6 m/s2 at an angle of 32.5° above the horizontal. If the ball lands at the same level from which it was kicked, for what amount of time was it in the air?
I'm stuck at the equation y = (v0 sin theta)t - 1/2 gt^2. I have y = (15.6*sin*theta)*t - 1/2 (9.81*t)^2. I keep seeing to put 0 in for y to solve for time, but I can't figure out for the life of me how to solve it from that point. Any help is appreciated.
1 Answer
- L. E. GantLv 73 years ago
You have a quadratic equation...
0 = 15.6*sin(32.5)* t - 4.905 * t^2
==> t(15.6*sin(32.5) - 4.905t)
==> t = 0 or t = 15.6 *sin(32.5)/4.905
