Preview Section 1 Introduction to Vectors Section 2 Vector Operations Section 3 Projectile Motion Section 4 Relative Motion

Projectile Motion Projectiles: objects that are launched into the air tennis balls, arrows, baseballs, wrestlers Gravity affects the motion Path is parabolic if air resistance is ignored Path is shortened under the effects of air resistance Discuss the wide variety of projectiles. Tell students that the effect of air resistance is significant in many cases, but we will consider ideal examples with gravity being the only force. The effects of air were not very significant in the coin demonstration (see the Notes on the previous slide), but would be much more significant if the objects were traveling faster or had more surface area. Use the PHET web site to allow students to study projectile motion qualitatively. http://phet-web.colorado.edu/web-pages/index.html Go to simulations, choose “motion,” and choose then choose “projectile motion.” In this simulation, you can raise or lower the canon. Start with horizontal launches and note that the time in the air is only dependent on the height, and not on the speed of launch. You can change objects, and you can even launch a car. You also have the option of adding air resistance in varying amounts, as well as changing the launch angle. Have students determine which launch angles produce the same horizontal distance or range (complimentary angles) and find out which launch angle gives the greatest range (45°). Ask them to investigate the effect of air resistance on these results.

Components of Projectile Motion As the runner launches herself (vi), she is moving in the x and y directions. Remind students that vi is the initial velocity, so it never changes. Students will learn in later slides that vx,i also does not change (there is no acceleration in the horizontal direction) but vy,i does change (because of the acceleration due to gravity).

Analysis of Projectile Motion Horizontal motion No horizontal acceleration Horizontal velocity (vx) is constant. How would the horizontal distance traveled change during successive time intervals of 0.1 s each? Horizontal motion of a projectile launched at an angle: Since the initial velocity is constant, the change in x for each successive time interval (such as 0.1 s) will always be the same. Point out that the ball moves the same distance sideways between successive time intervals. Many students mistakenly believe that the ball is falling straight down eventually. In fact, it keeps moving sideways at a steady rate in the absence of air resistance. With air resistance, it can eventually reach a point where it is falling nearly straight down.

Analysis of Projectile Motion Vertical motion is simple free fall. Acceleration (ag) is a constant -9.81 m/s2 . Vertical velocity changes. How would the vertical distance traveled change during successive time intervals of 0.1 seconds each? Vertical motion of a projectile launched at an angle: Students should note that the vertical distance increases during each successive time interval. The equations above are simply equations (2), (5), and (4) from the previous section. You might want to write the “old” equations on the board prior to showing them these “new” equations.

Projectile Motion Click below to watch the Visual Concept.

Projectile Motion - Special Case Initial velocity is horizontal only (vi,y = 0). Point out that these equations are the same as those on the previous slides with vi,y = 0 or a launch angle  = 0. These equations could be used for the coin as it fell off the table (see the Notes on the first slide of this section) or for an object dropped from an airplane flying at a level altitude. The previous equations (last two slides) are more general and apply to any projectile.

Projectile Motion – Launched at an angle When projectiles are launched at an angle we can use the following equations and use acceleration of gravity. ΔX= Vi (cosθ)Δt Vx = Vi (cosθ) = constant Point out that these equations are the same as those on the previous slides with vi,y = 0 or a launch angle  = 0. These equations could be used for the coin as it fell off the table (see the Notes on the first slide of this section) or for an object dropped from an airplane flying at a level altitude. The previous equations (last two slides) are more general and apply to any projectile.

Projectile Motion Summary Projectile motion is free fall with an initial horizontal speed. Vertical and horizontal motion are independent of each other. Horizontally the velocity is constant. Vertically the acceleration is constant (-9.81 m/s2 ). Components are used to solve for vertical and horizontal quantities. Time is the same for both vertical and horizontal motion. Velocity at the peak is purely horizontal (vy = 0). The 4th and 5th summary points are essential for problem solving. Emphasize these points now, and return to them as students work through problems.

Classroom Practice Problem (Horizontal Launch) A pelican flying along a horizontal path drops a fish from a height of 5.4m while traveling 5.0m/s. how far does the fish travel horizontally before it hits the water below? Answer: 5.0m As the students look at the equations, they will not find a single equation that allows them to solve this problem. First, as is often the case, they must solve for time using the height of the building (y) and the acceleration of gravity (ag). Then, they can use this time with the horizontal distance (x) to find the horizontal speed (vx).

Classroom Practice Problem (Horizontal Launch) Give both the horizontal and vertical components of the velocity of the fish from the previous problem before the fish enters the water. Answer: 5.0m/s, 9.8m/s As the students look at the equations, they will not find a single equation that allows them to solve this problem. First, as is often the case, they must solve for time using the height of the building (y) and the acceleration of gravity (ag). Then, they can use this time with the horizontal distance (x) to find the horizontal speed (vx).

Classroom Practice Problem (Horizontal Launch) A movie director is shooting a scene that involves dropping a stunt dummy out of an airplane and into a swimming pool. The plane is 10.0 m above the ground, traveling at a velocity of 22.5 m/s in the positive x direction . The director wants to know where the plane’s path the dummy should be dropped so that it will land in the pool? Answer: 32.2m As the students look at the equations, they will not find a single equation that allows them to solve this problem. First, as is often the case, they must solve for time using the height of the building (y) and the acceleration of gravity (ag). Then, they can use this time with the horizontal distance (x) to find the horizontal speed (vx).

Classroom Practice Problem (Horizontal Launch) Find the instantaneous velocity of the stunt dummy in the previous problem as it hits the water? Answer: 26.5 m/s at 31.913° below the horizontal As the students look at the equations, they will not find a single equation that allows them to solve this problem. First, as is often the case, they must solve for time using the height of the building (y) and the acceleration of gravity (ag). Then, they can use this time with the horizontal distance (x) to find the horizontal speed (vx).

Classroom Practice Problem (Horizontal Launch) A cat chases a mouse across a 1.0m high table. The mouse steps out of the way and the cat slides off the table at a speed of 5.0m/s. Where does the cat strike the floor? Answer: 2.2m from the table As the students look at the equations, they will not find a single equation that allows them to solve this problem. First, as is often the case, they must solve for time using the height of the building (y) and the acceleration of gravity (ag). Then, they can use this time with the horizontal distance (x) to find the horizontal speed (vx).

Classroom Practice Problem (Projectile Launched at an Angle) In a scene in an action movie, a stuntman jumps from the top of one building to the top of another building 4.0m away. After a running start, he leaps at an angle of 15° with respect to the flat roof while traveling at a speed of 5.0m/s. Will he make it to the other roof, which is 2.5 m shorter than the building he jumps from? Answer: Yes, -2.3 m One option is to first solve for t in the vertical motion equations. This requires the use of the quadratic equation. Then, t can be used to find the horizontal distance in the horizontal motion equations. The problem can also be divided into two parts and solved without a quadratic equation. First, find the time required to reach the peak where vy is zero. Then, find the height reached and add it onto the 15 m. Finally, find the time required to fall from this height, and use the total time to find the horizontal distance.

Classroom Practice Problem (Projectile Launched at an Angle) A golfer can hit a golf ball a horizontal distance of over 300m on a good day. What maximum height would a 301.5m drive reach if it were launched at an angle of 25.0° to the ground? (At the top of its flight, the ball’s vertical velocity will be zero) Answer: 70.3m One option is to first solve for t in the vertical motion equations. This requires the use of the quadratic equation. Then, t can be used to find the horizontal distance in the horizontal motion equations. The problem can also be divided into two parts and solved without a quadratic equation. First, find the time required to reach the peak where vy is zero. Then, find the height reached and add it onto the 15 m. Finally, find the time required to fall from this height, and use the total time to find the horizontal distance.

Classroom Practice Problem (Projectile Launched at an Angle) Salmon often jump waterfalls to reach their breeding grounds. Starting 2.00m from a waterfall 0.55m in height, at what minimum speed must a salmon jumping at an angle of 32.0° leave the water to continue upstream? Answer: 6.2 m/s One option is to first solve for t in the vertical motion equations. This requires the use of the quadratic equation. Then, t can be used to find the horizontal distance in the horizontal motion equations. The problem can also be divided into two parts and solved without a quadratic equation. First, find the time required to reach the peak where vy is zero. Then, find the height reached and add it onto the 15 m. Finally, find the time required to fall from this height, and use the total time to find the horizontal distance.

Classroom Practice Problem (Projectile Launched at an Angle) A quarterback throws the football to a receiver who is 31.5m down the field. If the football is thrown at an initial angle of 40.0 ° to the ground, at what initial speed must the quarterback throw the ball? What is the ball’s highest point during its flight? Answer: 17.7 m/s, 6.60m One option is to first solve for t in the vertical motion equations. This requires the use of the quadratic equation. Then, t can be used to find the horizontal distance in the horizontal motion equations. The problem can also be divided into two parts and solved without a quadratic equation. First, find the time required to reach the peak where vy is zero. Then, find the height reached and add it onto the 15 m. Finally, find the time required to fall from this height, and use the total time to find the horizontal distance.