Vocabulary Preview. Oct 21 9:53 AM. Projectile Motion. An object shot through the air is called a projectile.

Transcription

1 Projectile Trajectory Range Launch angle Vocabulary Preview Projectile Motion Projectile Motion An object shot through the air is called a projectile. A projectile can be a football, a bullet, or a drop of water. You can draw a free body diagram of a launched projectile and identify all the forces that are acting on it. No matter what the object is, after a projectile has been given an initial thrust, if you ignore air resistance, it moves through the air only under the force of gravity. The force of gravity is what causes the object to curve downward in a parabolic flight path. Its path through space is called its trajectory. 1

2 Projectile Motion Demonstration Remember the demonstration of free fall vs. projectile motion. Which ball hit the floor first? Projectile Motion Independence of Motion in Two Dimensions Click image to view movie. 2

3 With your partner Write a 10 word summary for each of the most 3 important points of the video clip. Projectiles launched horizontally If an object is fired horizontally, what are its velocity components? Why? v x = v y = Oct 6 8:12 AM 3

4 A projectile is launched horizontally at a velocity of 20 m/s from a height of 10 m above the ground. a) What is the time the projectile spends in the air? b) How far does the projectile move horizontally? c) What is the projectile's velocity (magnitude and direction) when it hits the ground? Oct 6 8:14 AM Δx v ox t Δy v oy v y g t b)? 20 m/s? 10 m 0 m/s? 9.8m/s 2 a)? a) time in air, t? Δy = v oy t + 1/2 g t 2 b) horizontal displacement, Δx =? 10 = 0 t + 1/2 ( 9.8) t 2 Δx = v ox t 2.04 = t 2 Δx =(20) (1.43) = 28.6 m t = 1.43 s c) velocity when hits the ground, v & θ? v x θ v x = v ox =20 m/s v 2 = v x 2 + v y 2 v 2 = (20) 2 + (14) 2 = 596 v = 24.4 m/s v v y v y = v oy + gt v y = 0 + ( 9.8)(1.43) = 14 m/s θ = tan 1 (v y /v x ) = = tan 1 (14 / 20) = 35 degrees below horizontal Oct 6 8:18 AM 4

5 Projectile Motion Projectiles Launched at an Angle When a projectile is launched at an angle, the initial velocity has a vertical component as well as a horizontal component. If the object is launched upward, like a ball tossed straight up in the air, it rises with slowing speed, reaches the top of its path, and descends with increasing speed. Check Question 1 A boy standing on a balcony drops one ball and throws another with an initial horizontal velocity of 3 m/s. Which of the following statements about the horizontal and vertical motions of the balls is correct? (Neglect air resistance.) The balls fall with a constant vertical velocity and a constant horizontal acceleration. The balls fall with a constant vertical velocity as well as a constant horizontal velocity. The balls fall with a constant vertical acceleration and a constant horizontal velocity. The balls fall with a constant vertical acceleration and an increasing horizontal velocity. 5

6 Check Answer 1 Answer: C Reason: The vertical and horizontal motions of a projectile are independent. The only force acting on the two balls is force due to gravity. Because it acts in the vertical direction, the balls accelerate in the vertical direction. The horizontal velocity remains constant throughout the flight of the balls. Check Question 2 Which of the following conditions is met when a projectile reaches its maximum height? Vertical component of the velocity is zero. Vertical component of the velocity is maximum. Horizontal component of the velocity is maximum. Acceleration in the vertical direction is zero. 6

7 Check Answer 2 Answer: A Reason: The maximum height is the height at which the object stops its upward motion and starts falling down, i.e. when the vertical component of the velocity becomes zero. Equations for Projectile Motion The equations for Projectile Motion are similar to those with 1 D motion. The biggest difference is to realize that there are two independent motions occurring simultaneously (horizontal & vertical) Because we now have velocities in the x and y directions, it is important for our labeling to clearly specify which we are concerned with. 7

8 Equations for Projectile Motion If you are given an initial velocity and an angle of launch, your first step should be to find the x & y components of the velocity. v xo = v o cos θ v yo = v o sin θ The x direction is the simpler of the two. There is nothing accelerating the projectile in the x direction therefore the motion can be summarized by: v xo = v x = constant v x = Δx / Δt Equations for Projectile Motion The y direction is free fall so those equations apply: Δy = v yo Δt + ½ g Δt 2 v y = v yo + g Δt 2 v y = v 2 yo + 2 g Δy Δy = ½ (v yo + v y ) Δt The only idea that links the two motions is time! 8

9 Practice Problem A ball is launched at 4.5 m/s at 66 above the horizontal. A) What is the maximum height? B) How long is the ball in the air? C) How far does the ball travel horizontally? Given: v o = 4.5 m/s θ = 66 g = -9.8 m/s 2 v xo = v o cos θ = 4.5 cos 66 = 1.83 m/s v yo = v o sin θ = 4.5 sin 66 = 4.11 m/s Unknown: y max t air Δx 9

10 a) At highest point, v yf = 0 m/s. 2 v y = v 2 yo + 2 g Δy 0 = (4.11) (-9.8) Δy = Δy Δy = 0.86 m b) The total time in the air is equal to the time up + the time down. Time up occurs when v y = 0 m/s v y = v yo + g Δt 0 = Δt = -9.8 Δt Δ tup = 0.42 s Δt total = 2 x Δ tup = 2(0.42)= 0.84 s 10

11 c) How far it travels horizontally is the x direction. v x = Δx / Δt 1.83 = Δx / 0.84 Δx = 1.54 m Oct 21 9:59 AM 11