 # Section 2 Extra Questions

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Section 2 Extra Questions
Preview Section 1 Projectile Motion Section 2 Extra Questions

Pythagorean Theorem and Tangent Function
Remind students that the Pythagorean theorem can only be used with right triangles.

Resolving Vectors Into Components
Review these trigonometry definitions with students to prepare for the next slide (resolving vectors into components).

Resolving Vectors into Components
Opposite of vector addition Vectors are resolved into x and y components For the vector shown at right, find the vector components vx (velocity in the x direction) and vy (velocity in the y direction). Assume that that the angle is 20.0˚. Answers: vx = 89 km/h vy = 32 km/h Review the first solution with students, and then let them solve for the second component.

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. 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 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 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)
People in movies often jump from buildings into pools. If a person jumps horizontally by running straight off a rooftop from a height of 30.0 m to a pool that is 5.0 m from the building, with what initial speed must the person jump? Answer: 2.0 m/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 (Projectile Launched at an Angle)
A golfer practices driving balls off a cliff and into the water below. The edge of the cliff is 15 m above the water. If the golf ball is launched at 51 m/s at an angle of 15°, how far does the ball travel horizontally before hitting the water? Answer: 1.7 x 102 m (170 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.

Now what do you think? Suppose two coins fall off of a table simultaneously. One coin falls straight downward. The other coin slides off the table horizontally and lands several meters from the base of the table. Which coin will strike the floor first? Explain your reasoning. Would your answer change if the second coin was moving so fast that it landed 50 m from the base of the table? Why or why not? By this point, students should have a solid understanding of the fact that both coins strike the floor at the same time (regardless of how fast the second coin is moving), and they should be able to explain why this is the case. You could also have students apply the projectile motion equations to the coin demonstration at this time.

Preview Multiple Choice Short Response Extended Response

Multiple Choice, continued
Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 9. What is the initial speed of the girl’s ball relative to the boy? A. 1.0 m/s C. 2.0 m/s B. 1.5 m/s D. 3.0 m/s

Multiple Choice, continued
Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 10. If air resistance is disregarded, which ball will hit the ground first? F. the boy’s ball H. neither G. the girl’s ball J. cannot be determined

Multiple Choice, continued
Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 10. If air resistance is disregarded, which ball will hit the ground first? F. the boy’s ball H. neither G. the girl’s ball J. cannot be determined

Multiple Choice, continued
Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 11. If air resistance is disregarded, which ball will have a greater speed (relative to the ground) when it hits the ground? A. the boy’s ball C. neither B. the girl’s ball D. cannot be determined

Multiple Choice, continued
Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 12. What is the speed of the girl’s ball when it hits the ground? F. 1.0 m/s H. 6.2 m/s G. 3.0 m/s J. 8.4 m/s

Short Response, continued
15. A ball is thrown straight upward and returns to the thrower’s hand after 3.00 s in the air. A second ball is thrown at an angle of 30.0° with the horizontal. At what speed must the second ball be thrown to reach the same height as the one thrown vertically?

Short Response, continued
15. A ball is thrown straight upward and returns to the thrower’s hand after 3.00 s in the air. A second ball is thrown at an angle of 30.0° with the horizontal. At what speed must the second ball be thrown to reach the same height as the one thrown vertically? Answer: 29.4 m/s

Extended Response 16. A human cannonball is shot out of a cannon at 45.0° to the horizontal with an initial speed of 25.0 m/s. A net is positioned at a horizontal distance of 50.0 m from the cannon. At what height above the cannon should the net be placed in order to catch the human cannonball? Show your work.

Extended Response 16. A human cannonball is shot out of a cannon at 45.0° to the horizontal with an initial speed of 25.0 m/s. A net is positioned at a horizontal distance of 50.0 m from the cannon. At what height above the cannon should the net be placed in order to catch the human cannonball? Show your work. Answer: 10.8 m