This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1 Lesson 1.4 Projectile Motion. 1. Components of motion. In most cases of objects in motion, the motion is not restricted to x or y directions alone. For example, a ball on a table given a sideways push moves as shown in the figure below. The horizontal push gives the ball an x velocity, let us call it v x . Since there is no pull or push in the horizontal direction, there is no acceleration in the xdirection. Therefore, this x velocity, namely v x , will remain constant throughout the fall of the ball. As soon as the ball leaves the table, it begins to fall because of the acceleration V V V V V x x x x v y =0 v y v y v y v y x Fig. 1. The horizontal velocity remains the same The vertical velocity increases due to gravity. Initially it has no downward or yvelocity. But with time, its downward velocity (v y ) increases as illustrated in the figure above. At any time, the ball has an xvelocity and a yvelocity. The path it takes is the result of an xmotion as well as a ymotion. These two motions are independent of each other. In other words, the vertical motion does not affect the horizontal motion. We can analyze one motion using the equations of motion as though the other does not exist. In this case let us analyze the x and y motions separately: (i) xmotion : The xvelocity, namely v x , remains the same throughout the fall. If the ball falls for a time t s, the horizontal displacement x can be obtained as x = v x t. (ii) ymotion : The yvelocity when t = 0 is 0. We will write it as v yo = 0. The velocity at any time v y is given by v y = v yo + at = 0 + (g) t = gt. The ydisplacement at any time can be written as: 2 y = y o + v yo t + (1/2)at 2 . y = y o – (1/2)g t 2 . Example 1: A boy throws a ball horizontally with a velocity of 10 m/s from the top of a building 50 m high. How far from the foot of the building will the ball fall? Solution: x x o =0 y =50 o Fig. 2 During the time it takes the ball to drop 50 m it travels a horizontal distance x We will consider the vertical and horizontal motions of the ball separately. For the vertical motion: y o = 50 m (We will take the origin at the ground level). v yo = 0 a = g = 9.8 m.s2 When the ball falls on the ground, y = 0 because we have the origin on the ground. If t is the time taken by the ball to fall to the ground, we can write: y = y o + v yo t + (1/2)at 2 . 0 = 50 + 0 + (1/2)(9.8) t 2 0 = 50 – 4.9 t 2 Solving this equation gives t = 3.2 s For the horizontal motion: During the 3.2 s it took the ball to hit the ground, it had a uniform horizontal velocity 10 m.s1 (v x = 10 m.s1 ). Therefore, the horizontal displacement x is given by: x = v x t = 10 ⋅ 3.2 = 32.0 m 3 Example 2 : A ball rolls off a table with a horizontal velocity of 2.0 m.s1 and falls 1.3 m away from the foot of the table. What is the height of the table?...
View
Full Document
 Fall '10
 George
 Physics, Equations, Projectile Motion, Velocity, ball

Click to edit the document details