PHYSICS 100A #85014

FALL 2003 QUIZ 2

1. (a ) Use components to find A + B – C as shown in figure. (b) Sketch the resultant in the figure.

   (a) We find the resultant from      Rx = Ax + Bx - Cx = (40) + (-30)cos45° - 0           = 18.8.      Ry = Ay + By - Cy = 0 + (30)sin45° - (-50)            = 71.2.      R = (Rx2 + Ry2)1/2          = [(18.8)2  + (71.2)2 ]1/2         = 73.6.      tanq = Ry/Rx= 71.2/18.8 = 3.7872, which gives q  = 75.4°(above +x-axis).

   (b)

2. A ball was thrown horizontally from the edge of the roof of 10.0-m-high building with a velocity of 5.0m/s. What was the ball’s velocity at 1.2s after it was thrown?

    Solution. From the problem we have v₀ = 5.00m/s, and q₀ = 0.  Choose the starting point as the origin with downward as positive.

                          v_0x =  v₀cosq₀ = 5m/s, and v_0y = 0.

                  At t =1.2s, we have

                          v_x = v_0x = 5m/s. 

                         v_y  = v_0y - gt = 0 - (9.80m/s²)(1.2s) =  -11.76m/s.

                 We find the velocity from

                         v = (v_x² + v_x²)^1/2 = [(5.0m/s)² + (-11.76m/s)²]^1/2 = 12.78m/s.

                         tanq = v_y/v_x = (-11.76m/s)/(5m/s) = -2.352, which gives q  = 67°.(below +x-axis)