Chapter 5: Exercises with Solutions

Section 5.5Motiont = −b ± √ b 2 − 4ac2at = −120 ± √ (120) 2 − 4(−16)(50)2(−16)t = −120 ± √ 14400 + 3200−32t = −120 ± √ 17600−32t ≈ −0.396, 7.896As usual, we throw out the negative time, so t ≈ 7.896 s. This agrees with the zerothat we found using the calculator.45. We are given that y 0 = 20 m and v 0 = 110 m/s, and the well-known accelerationdue to gravity is a = −9.8 m/s 2 when using the metric system. Plug these into theheight formula:y = y 0 + v 0 t + 1 2 at2y = 20 + 110t − 4.9t 2y = −4.9t 2 + 110t + 20This is a quadratic function and we can graph it. We enter it into our calculator usingx’s instead of t’s. A good window to use is Xmin = 0, Xmax = 30, Y min = −200and Y max = 1000. We only need to look at time greater than or equal to zero.To find when the ball hits the ground (the zero of the function) with your calculator,press 2nd TRACE to access the CALC menu and choose 2:zero.Use the left arrowto move the cursoralong the curve untilit is to the left of thefirst zero. Hit ENTER.Use the right arrow tomove the cursor until itis to the right of thesame zero. Hit ENTER.Finally hit ENTERnear that same zerofor the guess, andyou get the zero.We get (22.629349, 0).Now sketch the graph on your paper.Version: Fall 2007