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Name: ___________________________ Period: _______________ Date: __________ AKIBOLA CHAPTER 3 NOTES TWO DIMENSIONAL MOTION AND VECTORS Sect. 3-2: Equations of Kinematics in 2-Dimensions Equations representing motion of an object on the x-axis _________________________________________ _________________________________________ _________________________________________ _________________________________________ Equations representing motion of an object on the y-axis _________________________________________ _________________________________________ _________________________________________ _________________________________________ The ‘x’ part of the motion occurs exactly as it would if the ‘y’ part did not occur at all, and vice versa. Sect. 3-3: Projectile Motion What is a projectile? Projectiles are objects ________________________________________________________________ _______________________. Free fall implies _____________________________________________ o ____________________________________________ ________________________________ dimensions. o Examples of projectiles include ______________________________________________________________________ ______________________________________________________________________ _________________________________________________________ _______________________________________________________________________. Trajectory is __________________________________________________________________________________ __________________________________________________________________________________ o We can ______________________________________________________________________ __________________________________________ and __________________________________________ If a ball is thrown horizontally and another ball is dropped-----they will both o ____________________________________________!!!!!!!! 1 Name: ___________________________ Period: _______________ Date: __________ AKIBOLA Solving Projectile Motion Problems You need to _________________________________________________________________ o 1st part _______________________________________ o 2nd Part _______________________________________ o Keep the _____________________________________________________________ Horizontal motion equations and constants Vertical motion equations and constants Calculating time of fall If a projectile is _____________________________________________________________________ _____________________________________________________________________ For such projectiles, ________________________________________________________________ _____________________________________________________________________ 2 Name: ___________________________ Period: _______________ Date: __________ AKIBOLA Example 1 People in movies often jump from buildings into pools. If a person jumps from the 10th floor (30.0 m) to a pool that is 5 m away from the building, with what initial horizontal velocity must the person jump? Step 1: Write down Givens and unknowns. o You will have to _______________________________________________________ ____________________________________ X-Component Y-Component Step 2: Write down equations that might solve it. We can use ______________________________________________________________ Your Y-equation ________________________ We then use _____________________________________________________________ Your X-equation _______________________ These are the two most common equations I will use in this section!!!! Step 3: Carry out the solution. Time for the jump Step 4: Carry out the solution. Using our time to calculate the initial horizontal velocity 3 Name: ___________________________ Period: _______________ Date: __________ AKIBOLA Practice Problems Below: (We shall do the first one together) #1 An autographed baseball rolls off of a 0.70m high desk. And strikes the floor 0.25m away from the desk. How fast did it roll off the table? #2 A cat chases a mouse across a 1.0 m high table. The mouse steps out of the way and the cat slides off the table and strikes the floor 2.2 m from the edge of the table. What was the cats speed when it slid off the table? 4 Name: ___________________________ Period: _______________ Date: __________ AKIBOLA Physics – Chapter 3.3 Projectile Worksheet (H/W) 1. A pelican flying along a horizontal path drops a fish from a height of 5.4 m. The fish travels 8 m horizontally before it hits the water below. What is the pelican’s speed? 2. If the pelican in question 1 was traveling at the same speed but was only 2.7 m above the water, how far would the fish travel horizontally before hitting the water below? 3. Determine the launch speed of a horizontally launched projectile that lands 26.3 meters from the base of a 19.3-meter high cliff. 5 Name: ___________________________ Period: _______________ Date: __________ AKIBOLA 4. A soccer ball is kicked horizontally at 15.8 m/s off the top of a field house and lands 33.9 meters from the base of the field house. Determine the height of the field house. 5. A ball is projected horizontally from the top of a 92.0-meter high cliff with an initial speed of 19.8 m/s. Determine: (a) the horizontal displacement, (b) the final speed the instant prior to hitting the ground. 6 Name: ___________________________ Period: _______________ Date: __________ AKIBOLA Projectiles launched at an angle: As shown here, projectiles launched at an angle have ________________________________ ______________________________________________ Suppose the initial velocity vector makes an angle θ with the horizontal The ___________________________________ can be used to find ____________________ ____________________________________________________ o _________________________ o _________________________ Practice: The Kickoff A placekicker kicks a football at an angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, determine the following: a) Maximum height that the ball attains. b) Time of flight between kickoff and landing c) The range ‘R’ of the projectile o How is this different from the problems you’ve solved so far? _________________________________________________________________________ o How will we get horizontal and vertical velocity? _________________________________________________________________________ Problem Sketch 7 Name: ___________________________ Period: _______________ Date: __________ AKIBOLA Conceptual Example: Two Ways to Throw a Stone From the top of a cliff, a person throws two stones. The stones have identical initial speeds, but stone 1 is thrown downward at some angle above the horizontal and stone 2 is thrown at the same angle below the horizontal. Neglecting air resistance, which stone, if either, strikes the water with greater velocity? Explain 8 Name: ___________________________ Period: _______________ Date: __________ AKIBOLA Projectiles launched at an angle worksheet - I Show your work for full credit!! 1. An object is launched with an initial velocity of 25.0 m/s at an angle of 30° above horizontal. a. Determine the components of the initial velocity. [answers: vox = 21.7 m/s, voy = 12.5 m/s] b. Use the correct component to determine the time that it takes the object to reach the top of its flight. [answers:1.27 s] c. Calculate the maximum vertical height that the object will reach. [answers:7.96 m] d. Calculate the horizontal distance that the object will travel. [answers:55.1 m] 9 Name: ___________________________ Period: _______________ Date: __________ AKIBOLA 2. A ball is launched with an initial horizontal velocity of 10.0 m/s. It takes 0.5 s for the ball to reach its maximum height. a. Determine the maximum horizontal distance that the ball will travel. [answers:10 m] b. Calculate the initial vertical velocity of the ball. [answers:4.91 m/s] 3. If the initial velocity, equals 50.00 m/s, find the maximum height and range for each of the launch angles listed in the table below. 10 Name: ___________________________ Period: _______________ Date: __________ AKIBOLA Projectiles launched at an angle worksheet - II Show your work for full credit!! 4. Measurements made in 1910 indicate that the common flea is an impressive jumper, given its size. Assume that a flea’s initial speed is 2.2 m/s, and that it leaps at an angle of 21° with respect to the horizontal. If the jump lasts 0.16 s, a. What is the magnitude of the flea’s horizontal displacement? b. How high does the flea jump? 5. A volleyball is spiked so that it has an initial velocity of directed downward at an angle of below the horizontal. What is the horizontal component of the ball's velocity when the opposing player fields the ball? 6. As a tennis ball is struck, it departs from the racket horizontally with a speed of . The ball hits the court at a horizontal distance of 19.6 m from the racket. How far above the court is the tennis ball when it leaves the racket? (notice that this is a horizontally launched projectile) 11 Name: ___________________________ Period: _______________ Date: __________ AKIBOLA 7. A skateboarder shoots off a ramp with a velocity of , directed at an angle of above the horizontal. The end of the ramp is 1.2 m above the ground. Let the x axis be parallel to the ground, the direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. a. How high above the ground is the highest point that the skateboarder reaches? b. When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp? 8. A golfer hits a shot to a green that is elevated 3.0 m above the point where the ball is struck. The ball leaves the club at a speed of at an angle of above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands. 12 Name: ___________________________ Period: _______________ Date: __________ AKIBOLA KEY 5. SSM REASONING The vertical component of the ball’s velocity v0 changes as the ball approaches the opposing player. It changes due to the acceleration of gravity. However, the horizontal component does not change, assuming that air resistance can be neglected. Hence, the horizontal component of the ball’s velocity when the opposing player fields the ball is the same as it was initially. SOLUTION Using trigonometry, we find that the horizontal component is b g v x v 0 cos 15 m / s cos 55 8.6 m / s 6. REASONING y v0 y t We will determine the ball’s displacement y in the vertical direction from 1 a t2 2 y (Equation 3.5b). We choose this equation because we know that v0 y 0.00 m / s (the ball initially travels horizontally) and that a y 9.80 m/s2 (assuming that upward is the +y direction). We do not know the value of the ball’s time of flight t. However, we can determine it from the motion in the horizontal or x direction. SOLUTION From Equation 3.5b we have that y v0 y t a y t 2 0.00 m / s t a y t 2 a y t 2 1 2 1 2 1 2 The motion in the horizontal direction occurs at a constant velocity of v0 x 28.0 m / s and the displacement in the horizontal direction is x 19.6 m . Thus, it follows that x v0 xt or t x 19.6 m 0.700 s v0 x 28.0 m / s Using this value for the time in the expression for y shows that 1 2 y a yt 2 1 2 9.80 m/s2 0.700 s 2 2.40 m Since the displacement of the ball is 2.40 m downward, the ball is 2.40 m above the court when it leaves the racket. 13 Name: ___________________________ Period: _______________ 7. Date: __________ AKIBOLA REASONING a. Here is a summary of the data for the skateboarder, using v0 = 6.6 m/s and = 58°: y-Direction Data y ? ay −9.80 m/s 2 vy v0y 0 m/s +(6.6 m/s)sin 58° = +5.6 m/s t x-Direction Data x ? ax vx v0x 0 m/s2 t +(6.6 m/s)cos 58° = +3.5 m/s We will use the relation v2y v02y 2a y y (Equation 3.6b) to find the skateboarder’s vertical displacement y above the end of the ramp. When the skateboarder is at the highest point, his vertical displacement y1 above the ground is equal to his initial height y0 plus his vertical displacement y above the end of the ramp: y1 = y0 + y. Next we will use v y v0 y a y t (Equation 3.3b) to calculate the time t from the launch to the highest point, which, with x v0 xt 12 axt 2 (Equation 3.5a), will give us his horizontal displacement x. SOLUTION a. Substituting vy = 0 m/s into Equation 3.6b and solving for y, we find 0 m/s v02y 2a y y 2 or y v02 y 2a y Therefore, the highest point y1 reached by y1 = y0 + y = 1.2 m + 1.6 m = +2.8 m above the ground. 5.6 m/s 2 2 9.80 m/s2 the 1.6 m skateboarder occurs when b. We now turn to the horizontal displacement, which requires that we first find the elapsed time t. Putting vy = 0 m/s into Equation 3.3b and solving for t yields v0 y (1) 0 m/s v0 y a y t or t ay Given that ax = 0 m/s2, the relation x v0 xt 12 axt 2 (Equation 3.5a) reduces to x v0 x t . Substituting Equation (1) for t then gives the skateboarder’s horizontal displacement: v0 y v0 x v0 y 3.5 m/s 5.6 m/s x v0 xt v0 x 2.0 m ay ay 9.80 m/s 14 Name: ___________________________ Period: _______________ Date: __________ AKIBOLA 8. REASONING The magnitude (or speed) v of the ball’s velocity is related to its x and y velocity components (vx and vy ) by the Pythagorean theorem; v v x2 v 2y (Equation 1.7). The horizontal component vx of the ball’s velocity never changes during the flight, since, in the absence of air resistance, there is no acceleration in the x direction (ax = 0 m/s2). Thus, vx is equal to the horizontal component v0x of the initial velocity, or vx v0 x v0 cos 40.0 . Since v0 is known, vx can be determined. The vertical component vy of the ball’s velocity does change during the flight, since the acceleration in the y direction is that due to gravity (ay = 9.80 m/s2). The relation v 2y v02 y 2a y y (Equation 2.6b) may be used to find v 2y , since ay, y, and v0y are known (v0y = v0 sin 40.0º). SOLUTION The speed v of the golf ball just before it lands is v v x2 v 2y v0 cos 40.0 2 v02 y 2a y y v 2y v0 cos 40.0 v0 sin 40.0 2 2 2a y y 14.0 m/s cos 40.0 14.0 m/s sin 40.0 2 9.80 m/s 2 3.00 m 11.7 m/s 2 2 15