Motion of a Projectile (Section 12.6) (2 of 2) For illustration, consider the two balls on the left. The red ball falls from rest, whereas the yellow ball is given a horizontal velocity. Each picture in this sequence is taken after the same time interval. Notice both balls are subjected to the same downward acceleration since they remain at the same elevation at any instant. Also, note that the horizontal distance between successive photos of the yellow ball is constant since the velocity in the horizontal direction is constant. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Kinematic Equations: Horizontal Motion Since constant a x  0, the velocity in the horizontal direction remains (v x  vox ) and the position in the x direction can be determined by: Why is x  x o  (vox ) t ax equal to zero (what assumption must be made if the movement is through the air)? Copyright © 2016 Pearson Education, Inc. All Rights Reserved Kinematic Equations: Vertical Motion Since the positive y-axis is directed upward, a y  – g. Application of the constant acceleration equations yields: v y  voy – g t y  yo  ( v oy ) t – ½ g t 2 v y 2  v oy 2 – 2 g ( y – y o ) For any given problem, only two of these three equations can be used. Why? Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example 12-12 (1 of 3) The chipping machine is designed to eject wood chips at v 0  25 ft s as shown in Figure 12-22. If the tube is oriented at 30° from the horizontal, determine how high, h, the chips strike the pile if at this instant they land on the pile 20 f t from the tube. ee Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example 12-12 (2 of 3) Solution Coordinate System. When the motion is analyzed between points O and A, the three unknowns are the height h, time of flight velocity v  A y tOA , and vertical component of . [Note that (v A )x = (v O )x . ] With the origin of coordinates at O, Figure 12–22, the initial velocity of a chip has components of (v O )x = (25 cos 30 ) ft s = 21.65 ft s  (v O )y = (25 sin 30 ) ft s = 12.5 ft s  Also, (v A )x = (v O )x = 21.65 ft s and a y = -32.2 ft s 2 . Since we do not need to determine v  A y , we have Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example 12-12 (3 of 3) Horizontal Motion.    xA = xO +  vO  x tOA 20 ft  0  (21.65ft s)tOA tOA = 0.9238 s Vertical Motion. Relating tOA to the initial and final elevations of a chip, we have    y A  yO  (vO ) y t OA  12 a c t 2 OA (h  4ft)  0  (12.5ft s)(0.9238 s) + 12 (32.3ft s 2 )(0.9238 s) 2 Answer h  1.81 ft NOTE: We can determine v  A y by using (v A ) y  (vO ) y  a c t OA . Copyright © 2016 Pearson Education, Inc. All Rights Reserved