Presentation on theme: "In this chapter you will: Use Newton’s laws and your knowledge of vectors to analyze motion in two dimensions. Solve problems dealing with projectile."— Presentation transcript:
In this chapter you will: Use Newton’s laws and your knowledge of vectors to analyze motion in two dimensions. Solve problems dealing with projectile and circular motion. Solve relative velocity.
SECTION 6.1 PROJECTILE MOTION Objectives Recognize that the vertical and horizontal motions of a projectile are independent. Relate the height, time in the air, and initial vertical velocity of a projectile using its vertical motion, and then determine the range using the horizontal motion. Explain how the trajectory of a projectile depends upon the frame of reference from which it is observed.
INTRO Parabola – set of all points equidistant from a fixed line called the directix, and a fixed point not on the line called the focus. Will be a U shaped graph. Projectile – an object shot through the air, such as a football, that has Independent Vertical and Horizontal motions and after receiving an initial thrust travels through the air only under the force of gravity. From Old Book, motion of objects given an initial velocity that then move only under the force of gravity. Trajectory – the path of a projectile through space.
INDEPENDENCE OF MOTION IN TWO DIMENSIONS Example of dropping a softball and launching one horizontally at 2 m/s. In both cases the horizontal acceleration is ZERO. (Dropped ball does not move horizontally and launched ball had a constant velocity thus no acceleration. (See Figure 6-1) Also in this example you see that the dropped and launched balls have the same vertical motion. Both balls are accelerated downward by the force of gravity. Both balls would hit the ground at the same time. Similar to the boat taking the same time to get across the river if there were no flow downstream and if there was a current downstream. The horizontal motion of the thrown ball does not affect its vertical motion at all. The horizontal and vertical components are Independent of each other.
INDEPENDENCE OF MOTION IN TWO DIMENSIONS The combination of a Constant Horizontal Velocity and Uniform Vertical Acceleration (Gravity) produces a Trajectory that has a Parabolic Shape. Since Horizontal and Vertical parts are Independent of each other if you find the time of one the other is the same, similar to the boat across the river problems. The shape of the trajectory and the horizontal motion depend on the viewpoint or frame of reference of the observer, But the Vertical Motion does not.
INDEPENDENCE OF MOTION IN TWO DIMENSIONS x = v x twhere x is the horizontal displacement, v x is the initial horizontal velocity, and t is the time v xf = v i where v xf is the final horizontal velocity and v i is the initial velocity y = v y t + ½ gt 2 where y is the vertical displacement, v y is the initial vertical velocity, t is time, and g is gravity v yf = v y + gtwhere v yf is the final vertical velocity and v y is initial vertical velocity Do Practice Problems p. 150 # 1-3
PROJECTILES LAUNCHED AT AN ANGLE When a projectile is launched at an angle the initial velocity has a vertical and horizontal component. Max Height – the height of the projectile when the vertical velocity is zero. The max height occurs when the time is HALF of the entire flight time (Example if time is 5 seconds then the time for the max height is 2.5 seconds). Range – denoted by R; the horizontal distance between the launch point of the projectile and where it returns to launch height; or the horizontal distance from the point of bounce until the projectile returns to the surface height. Range is the horizontal distance traveled during the entire flight time.
PROJECTILES LAUNCHED AT AN ANGLE v x = v i cos orA x = A i cos v y = v i sin orA y = A i sin
PROJECTILES LAUNCHED AT AN ANGLE Do Example 1 p. 151 v x = v i cos = 4.5(cos 66) = 4.5(.407) = 1.83 m/s v y = v i sin = 4.5(sin 66) = 4.5(.914) = 4.11 m/s A) Timeb) Max Height y = v y t + ½ gt 2 Since the trajectory is symmetric 0 = v y t + ½ gt 2 the max height occurred at.4165 s -4.11t = ½ (-9.8)t 2 So y = v y t + ½ gt2 t t y = 4.11(.4195) + ½ (-9.8)(.4195) 2 -4.11 = -4.9t y = 1.724 + (-4.9)(.176) -4.11 / -4.9 = t y = 1.724 -.862.839 s = t y =.862 m C) On next slide
PROJECTILES LAUNCHED AT AN ANGLE c) Range is the horizontal distance traveled during the entire flight time. So R = v x t R = 1.83(.839) R = 1.535 m Do Practice Problems p. 152 # 4-6
TRAJECTORIES DEPEND UPON THE VIEWER Remember that the force due to air resistance exists and it can be important but for now we are ignoring it. Do 6.1 Section Review p. 152 # 7-11