Chapter 5: Exercises with Solutions

Section 5.5Motionx = x 0 + v 0 t + 1 2 at20 = 5 m + (100 m/s)t + ( 1 2 )(−9.8m s 2 )t20 = 5 m + (100 m/s)t + (−4.9 m s 2 )t20 = (−4.9 m/s 2 )t 2 + (100 m/s)t + (5) mThis is a quadratic equation with a = −4.9 m/s 2 , b = 100 m/s and c = 5 m, so use thequadratic formula to solve.t = −b ± √ b 2 − 4ac2at = −100 m/s ± √ (100 m/s) 2 − 4(−4.9 m/s 2 )(5 m)2(−4.9 m/s 2 )t = −100 m/s ± √ 10000 m 2 /s 2 + 98 m 2 /s 2−9.8 m/s 2t = −100 m/s ± √ 10098 m 2 /s 2−9.8 m/s 2t ≈t ≈t ≈−100 m/s ± 100.49 m/s−9.8 m/s 2(−100 ± 100.49)m/s(−9.8)m/s 2−100 ± 100.49−9.8× m/sm/s 2−100 ± 100.49t ≈ × m −9.8 s × s2 m−100 ± 100.49t ≈ × m −9.8 s × s × smt ≈ 20.46 s, −0.50 sWe throw out the negative value for time. Thus the ball will hit the ground in approximatelyt = 20.46 s.43. We are given that y 0 = 50 ft and v 0 = 120 ft/s, and the well-known accelerationdue to gravity is a = −32 ft/s 2 when using ft. Plug these into the height formula:y = y 0 + v 0 t + 1 2 at2y = 50 + 120t − 16t 2y = −16t 2 + 120t + 50Version: Fall 2007