CHAPTER 2 Giancoli: Physics

t = [ 9 +/- SQRT( 92. + 4 X 4.9 X 2.2) ] / 9.8. The solutions of this quadratic equation are t = – 0.22 s, 2.06 s. Because the shot is released at t = 0, the physical answer is 2.06 s. We find the horizontal distance from. x = v0xt; x = (14 m/s)(cos 40°)(2.06 s) = 22 m. 30. (a) Because the athlete lands at the same level, we can use the ... ................
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