| Quiz 3 - Chapter 3 |
| Simulations Projectile Motion |
Ch. 3: 9, 10, 11, 12, 17, 18, 20, 22, 24, 31, 63, 64 Click here for the correction to the hint to #68, Quiz 1. Scroll down to #68. |
| turn in: 10, 12, 18, 20, 22, 24, 64 |
| 9. (a) See step 4. of PROBLEM SOLVING box on page
53. note you are using the realted angle so that Vx = -(735)cos(90 - 41.5) (b) D = (735 km/h)( 3.00 h) = length of displacement vector. Find -Dcos(90 - 41.5) and Dsin(90 - 41.5) |
| 10. See Example 3-3.Lay the vectors tip to tail as
in that example. . Compute each compnents then add the
compnes o get Dx =Ax + Bx + Cx Dy =Ay + By + Cy. For example: Ax = 44cos 28 and Ay = 44sin28 and Bx = -(26.5)cos56 and By = (26.5)sin56 |
| 11. Now subtract the components: Dx = Ax - Cx, where Cx = 0. Note: Cy = -31.0 for Dy = Ay - Cy |
| 12. (a) Now subtract the components: Dx = Bx - Ax and Dy = By - Ay (b) Now subtract the components: Dx = Ax - Bx and Dy = Ay - By |
| 17. Note: change in x = (3.5 m/s)t and find t from 6.5 m = (1/2)gt2. |
| 18. Note: change in x = Vxt and find H = (1/2)gt2. |
| 20. 5.0 m = Vfxt where t is found from Vfy = 0 = V1y - gt. But Vfy2 =0 = V1y2 - 2g(4.5 m): from this equation find V1y . Then plug into the first equation to get t. |
| 22. See example 3.8, where the total time is given in the derivation of the range R. |
| 24. (a) Use formula for range R in example 3.8 to find Vo. (b) Re-evaluate the Range R by using (1 + 0.05)Vo instead of Vo . |
| 31. (a) and (b) See example 3.9. (c) See Table 3.2. Use equation 2.11 (a) for the final components Vx and Vy. See example 3.5 for a lesson on finding the initial velocity componentsV1x , V1y . (d) V2 = Vx2 and Vy2. Find V by taking positive square root. (e) Use tan Ø = | Vy /|Vx| to get the magnitude of the angle below the x-axis. (f) Vfy2 =0 = V1y2 - 2g(change in y), where V1y is the initial y-component of velocity . See example 3.5 on how to compute it. Solve for (the change in y). |
| more discussions later... |