Answer to Chapter 3

1. The resultant is  31m, 44° N of E.
2. (a) V1x = -8.08,   V1y =  0;

             V2x = V2cos45° = 4.51cos45° = 3.19,

             V2y = V2sin45° = 4.51sin45° = 3.19.

        (b) For the components of the resultant we have

                  Rx = V1x + V2x = -8.08 + 3.19 = -4.89;

                  Ry = V1y + V2y = 0 + 3.19 = 3.19.

             We find the resultant from

                  R = (Rx2 + Ry2)1/2 = [(-4.89)2 + (3.19)2]1/2 = 5.84;

                  tan q = Ry/Rx = (3.19)/(4.89) = 0.652, which gives

                  q = 33.1° above  -x axis.

            Note that we have used the magnitude of Rx for the angle indicated on the diagram

  
3. (a) For the components we have

                    Rx = Cx - Ax - Bx

                       = 0 - 66.0cos28.0° - (-40.0cos56.0°)

                       = -35.9;

                    Ry = Cy - Ay - By

                       = -46.8 -66.0sin28.0° - 40.0sin56.0°

                       =  -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(66.0cos28.0°) - 3(-40.0cos56.0°) + 2(0)

                          = 183.8;

                       Ry = 2Ay - 3By + 2Cy

                            = 2(66.0sin28.0°) - 3(40.0sin56.0°) + 2(-46.8)

                            = -131.2.

                     We find the resultant from

                         R = (Rx2 + Ry2)1/2 = [(183.8)2 + (-131.2)2]1/2

                            =  266;

                         tan q = Ry/Rx =(131.2)/(183.8) = 0.714,

                      which gives 

                         q =35.5° below +x axis.                        

                       

  
4. The horizontal velocity is constant, and the vertical velocity will be zero when the pebbles hit the window. Using the coordinate system shown,    we find the vertical component of the initial velocity from

                  vy2 = v0y2 + 2a(h - y0)

                   0 = v0y2 + 2(-9.80m/s2)(8.0m - 0), which gives v = 12.5m/s. which gives v0y = 12.5m/s.

         (We choose the positive square root because we know that the pebbles are thrown upward.)

  
         We find the time for the pebbles to hit the window from the vertical motion.

                  vy = v0y + at;

                  0 = 12.5m/s + (9.80m/s2)t, which gives   t = 1.28s.

          For the horizontal motion we have

                   x = x0 + v0xt,;

                   9.0m = 0 + v0x(1.28s), which gives v0x = 7.0m/s.

          Because the pebbles are traveling horizontally when they hit the window, this is their speed.

5. (a) At the highest point, the vertical velocity vy = 0. We find the maximum height h from

                     vy2 = v0y2 + 2ay(y - y0);

                    0 = [(75.2m/s)sin34.5°]2 + 2(-9.80m/s2)(h - 0), which gives   h = 92.6m.

           (b) Because the projectile returns to the same elevation, we have

                      y = y0 + v0yt + 1/2ayt2;

                      0 = 0 + (75.2m/s)(sin34.5°)t + 1/2(-9.80m/s2)t2, which give t = 0, and 8.69s.

                 Because t = 0 was the launch time, the total time in the air was   8.69s.

           (c) We find the horizontal distance from

                       x = v0xt = (75.2m/s)(cos34.5°)(8.69s) =  539m.

           (d) The horizontal velocity will be constant: vx = vx0 = (75.2m/s)cos34.5° = 62.0m/s.

                 We find the vertical velocity from

                       vy = v0y + ayt = (75.2m/s)sin34.5° + (-9.80m/s2)(1.50s) = 27.9m/s.

                  The magnitude of the velocity is

                       v = (vx2 + vy2)1/2 = [(62.0m/s)2 + (27.9m/s)2]1/2 =  68.0m/s.

                  We find the angle from

                        tan q = vy/vx = (27.9m/s)/(62.0m/s) = 0.450, which gives   q =24.2° above the horizontal.

 

SEND QUESTIONS BACK TO PROBLEMS