ࡱ> `b_M $bjbj== "TWW\l8<"tC/@}}}V.X.X.X.X.X.X.$0 2|.}}}}}|..}V.}V.B ",h. 5%>-.D/0C/N-)3)3.CHAPTER 2,3 Giancoli: Physics Study Guide Dr. Lee Summary for FREE FALL: Key Concepts and equations: Take the upward direction to be positive. g means 9.8 m/s2. Use the constant acceleration equation with a = - g. v = vo - g t (1) The height covered, y, is given by: y = vo t - g t2 (2) These two equations is the key. They can be used to solve all problems A third equation is helpful but not necessary: v2 = vo2 - 2 g y Solution of some problems. (Note: capital X means multiplication) 35. We use a coordinate system with the origin at the top of the building and up positive. (a) To find the time of fall, we have eqn (2) with y = - 380 and vo = 0. - 380 = - (9.80 m/s2)t2 , which gives t = 8.81 s. (b) We find the velocity just before landing from v = - g t = - (9.80 m/s2)(8.81 s) = - 86.3 m/s. (The negative sign means that velocity is downward.) 37. We use a coordinate system with the origin at the ground and up positive. We use equations (1) and (2): v = vo - g t (1) y = vo t - g t2 (2) Let T be the time it took for the kangaroo to reach the top. From (2) we have: 2.7 = vo T - 9.8 T2 (3) We cannot find T yet because we do not know vo . To get that we use (1) knowing that at time T, v = 0. From (1) we have 0 = vo - 9.8 T vo = 9.8 T Putting this into eqn (3) gives : 2.7 = 9.8 T2 - 9.8 T2 = 9.8 T2 Therefore T = 0.74 s It takes the same time to go up and to go down. So the kangaroo is in the air for 1.5 s. 38. We use a coordinate system with the origin at the ground and up positive. Again we use eqn (1) and (2) v = vo - g t (1) y = vo t - g t2 (2) When the ball reaches the top, its speed is 0 and t = 3.3 / 2 = 1.65 s. Putting these into (1): 0 = vo - 9.8 X 1.65 Therefore, vo = 9.8 X 1.65 = 16.2 m/s To find the height reached, use eqn (2): Height reached = y = 16.2 X 1.65 - - 9.8 X (1.65)2 = 12.9 m  41. We use a coordinate system with the origin at 105 m and up positive. When the parcel is dropped, what is its speed ? It is NOT 0, it has the speed of the helicopter, + 5.5 m/s. This is the vo . Again we only need eqn (2). At the time T when it reaches ground, y = - 105 Therefore: - 105 = +5.5 T - 9.8 T2 This is a quadratic equation for T2 . We write it in the standard form: 4.9 T2 - 5.5 T 105 = 0 Using the quadratic formula gives: T = ( 5.5 +/- SQRT(5.52 + 4 X 4.9 X 105) /9.8 = - 4.1 s and 5.22 s We take the positive answer. The time to reach ground is 5.22 s. CHAPTER 3 Key Concepts and equations for section 1 to 4. Scalers have magnitude only. Examples are mass, temperature. Vectors have magnitude and direction. Examples are velocity, displacement, force. Resolving a vector into its components. If vector A makes an angle  with the x-axis, we can describe the vector by its x component of A = Ax = A cos  y component of B = Ay = A sin  Vector Addition ( A + B = C ) Graphical Method: When vector A is added to vector B, we can visualize it as going along A first. When we have reached the end of A, we follow along B. The resultant is a vector that runs from the beginning of A to the end of B. Algebraic Method: Resolve A and B into their components. The Resultant C has x component = Cx = Ax + Bx The Resultant C has y component = Cy = Ay + By Solution of some problems.  5. Draw the three vectors in scale as above The resultant is 31 m, 44 N of E. 11.  (a) For the components we have Rx = Ax + Bx + Cx = 66.0 cos 28.0 40.0 cos 56.0 + 0 = 35.9; Ry = Ay + By + Cy = 66.0 sin 28.0 + 40.0 sin 56.0  46.8 = 17.3. (b) We find the resultant from R = (Rx2 + Ry2)1/2 = [(35.9)2 + (17.3)2]1/2 = 39.9; tan q = Ry/Rx = (17.3)/(35.9) = 0.483, which gives q = 25.8 above + x-axis. 12. For the components we have  Rx = Ax  Cx = 66.0 cos 28.0  0 = 58.3; Ry = Ay  Cy = 66.0 sin 28.0  ( 46.8) = 77.8. We find the resultant from R = (Rx2 + Ry2)1/2 = [(58.3)2 + (77.8)2]1/2 = 97.2; tan q = Ry/Rx = (77.8)/(58.3) = 1.33, which gives q = 53.1 above + x-axis. 15. (a) For the components we have  Rx = Cx Ax Bx = 0 66.0 cos 28.0 ( 40.0 cos 56.0) = 35.9; Ry = Cy Ay By = 46.8 66.0 sin 28.0 40.0 sin 56.0 = 111.0. We find the resultant from R = (Rx2 + Ry2)1/2 = [( 35.9)2 + ( 111.0)2]1/2 = 117; tan q = Ry/Rx = (111.0)/(35.9) = 3.09, which gives q = 72.1 below  x-axis. (b) For the components we have Rx = 2Ax  3Bx + 2Cx = 2(66.0 cos 28.0)  3( 40.0 cos 56.0) + 2(0) = 183.8; Ry = 2Ay  3By + 2Cy = 2(66.0 sin 28.0 )  3(40.0 sin 56.0) + 2( 46.8) =  131.2. We find the resultant from R = (Rx2 + Ry2)1/2 = [(183.8)2 + ( 131.2)2]1/2 = 226; tan q = Ry/Rx = (131.2)/(183.8) = 0.714, which gives q = 35.5 below + x-axis. 20. We choose a coordinate system with the origin at the takeoff point, with x horizontal and y vertical, with the positive direction up. We find the height of the cliff from the vertical displacement: y = v0yt + ayt2; y = 0 + (-9.80 m/s2)(3.0 s)2 = - 44 m. - 44 m means that it its 44 m beneath the takeoff point. The horizontal motion will have constant velocity. We find the distance from the base of the cliff from x = v0xt; x = (1.6 m/s)(3.0 s) = 4.8 m.  24. We choose a coordinate system with the origin at the release point, with x horizontal and y vertical, with the positive direction up. We find the time of fall from the vertical displacement: y = ayt2; -56 m = (-9.80 m/s2)t2, which gives t = 3.38 s. The horizontal motion will have constant velocity. We find the initial speed from x = v0xt; 45 m = v0(3.38 s), which gives v0 = 13 m/s. Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 2 Page 2 >VEFQR$%&')=>?@AGLMmnxy) * 7 8 q r $ % 2 3 @ A T U _ `   ? 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