This equation can easily be factored and solved as follows: (b) To find the time it takes for the ball to return to the ground we can proceed as follows: (a) At the maximum height, the velocity is horizontal, so v y = 0 and this occurs at time The components of the initial velocity are as follows: SOLUTION We take the y direction as positive upward. Assume the ball leaves the foot at ground level, and ignore air resistance.
Calculate (a) the maximum height, (b) the time of travel before the football hits the ground, (c) how far away it hits the ground, (d) the velocity vector at the maximum height, and (e) the acceleration vector at the maximum height. Thus the acceleration of the object is that due to gravity, which acts downward with magnitude g = 9.80 m/s 2, and we assume it is constant.Ī football is kicked at an angle θ 0 = 37.0 o to the ground and with a velocity of 20.0 m/s. We consider only its motion after it has been projected and is moving freely through the air under the action of gravity alone. We will not be concerned now with the process by which the object id thrown or projected. Although air resistance is often importent, in many cases its effect can be ignored, and we will ignore it in the following analysis. These are all examples of projectile motion, which we can describe as taking place in two dimensions. Now we examine the more general motion of objects moving through the air in two dimensions near the surface of the Earth, such as a gold ball, a thrown or batted baseball, kicked footballs, speeding bullets, and athletes doing the long jump or high jump. On the last webpage we studied the motion of objects in one dimension in terms of displacement, velocity, and acceleration, including purely vertical motion of falling bodies undergoing acceleration due to gravity. The negative sign means θ = 38.5 o below the x axis. We can also specify the resultant vector by giving its magnitude and angle as follows: This specifies the resultant vector completely: Notice that D 2 y is negative because this vector points along the negative y axis. Since D 1 has magnitude 22.0 km and points north, it only has a y component: Let D 2 be the displacement vector from the first town to the second town, which is, as was stated, 60.0 o south of east. Let D 1 be the displacement vector from the origin to the first town pointing up the y axis, which is north. We choose the positive x axis to be east and the positive y axis north, and resolve each displacement vector into its components. SOLUTION We want to find her resultant displacement from the origin. What is her displacement from the post office? She then drives in a direction 60.0 o south of east for 47.0 km to another town. A rural mail carrier leaves the post office and drives 22.0 km in a northerly direction to the next town.
0 kommentar(er)
