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3.2 Equations of Kinematics in Two Dimensions Equations of Kinematics

3.2 Equations of Kinematics in Two Dimensions – the horizontal component

3.2 Equations of Kinematics in Two Dimensions – the vertical component

3.2 Equations of Kinematics in Two Dimensions The x part of the motion occurs exactly as it would if the y part did not occur at all, and vice versa.

Projectile Motion Chapter 3 Section 3

What is a projectile? When a long jumper approaches his jump, he runs along a straight line which can be called the x-axis. When he jumps, as shown in the diagram below, his velocity has both horizontal and vertical components

What is a projectile? Projectiles are objects in free fall that have an initial horizontal velocity. Free fall implies that the only force acting on the object is gravity. Initial horizontal velocity results in motion in two dimensions. Examples of projectiles include An arrow flying towards a target A baseball thrown in the air at an angle other than 90 o, unless they are thrown straight up.

Projectile motion Projectiles follow parabolic trajectories. Trajectory is the path of a flying object. The horizontal and vertical motions are independent. We can write separate equations of motion for each direction. Horizontal motion on the x-axis and the vertical motion on the y-axis

What is a projectile? If a ball is thrown horizontally, (yellow) and another ball is dropped (red)-----they will both HIT THE GROUND AT THE SAME TIME!!!!!!!! We shall do a demo of this

Solving Projectile Motion Problems You need to separate the motion down into 2 parts! The first part is the horizontal movement The second part is the vertical movement Keep the two sets of information separate using a table when solving such problems

The equations of motion for a projectile The change in displacement will now be identified as either: change in displacement on the X-axis (∆x) change in displacement on the Y-axis (∆y) or

Horizontal motion equations and constants

Vertical motion equations and constants

Calculating time of fall If a projectile is launched horizontally, this tells us that the object has an initial horizontal velocity (ie: in the x-axis) For such projectiles, the initial vertical velocity is zero (ie. in the y-axis) (v oy =0)

Example 1 People in movies often jump from buildings into pools. If a person jumps from the 10 th 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. You will have to separate your y-components from your x-components using a table X- ComponentY - Component  x = 5.0 m (from building to pool)  y = m (negative due to downward direction of fall) v ox = ?v oy = 0 (horizontal projectile) a x = 0a y = m/s 2

Step 2: Write down equations that might solve it. We can use the y equation to solve for the time of the jump. We then use this calculated time in the x equation to solve for the initial velocity. These are the two common equations you will use in this section!!!

Step 3: Carry out the solution. Time for the jump Using the y equation to solve for the time of the jump.

Step 4: Carry out the solution. Using our time to calculate the initial horizontal velocity Using the x equation to solve for the initial horizontal velocity.

Practice Problem 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?

Practice Problem # 1 Step 1: Write down what’s Given and your unknowns. (remember to separate your X and Y components using a table) X- ComponentY - Component  x = 0.25 m (from base of desk)  y = -0.7 m (negative due to downward direction of fall) v ox = ?v oy = 0 (horizontal projectile) a x = 0a y = m/s 2

Step 2: Write down equations that might solve it. We can use the y equation to solve for the time of the jump. We then use the time in the x equation to solve for the initial velocity.

Step 3: Carry out the solution. Time for the jump Using the y equation to solve for the time of the jump.

Step 3: Carry out the solution. Using our time to calculate the initial horizontal velocity Using the x equation to solve for the initial velocity.

Practice Problem # 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?

Practice Problem # 2 Step 1:  y = -1.0 m,  x = 2.2 m, v oy = 0 a x = 0, a y = -g v ox = ? Step 2:  y = v oy  t + ½ a y  t 2  x = v ox  t

Practice Problem # 2

Let us review the concepts for Horizontally launched projectiles

3.3 Projectile Motion The airplane is moving horizontally with a constant velocity of +115 m/s at an altitude of 1050m. a)Determine the time required for the Care package to hit the ground. b)What are the magnitude and direction of the final velocity of the Care package? Practice Problem # 2 - A Falling Care Package

Practice Problem # 1 Step 1: Write down what’s Given and your unknowns. (remember to separate your X and Y components using a table) X- ComponentY - Component  x =  y = m (negative due to downward direction of fall) v ox = 115 m/sv oy = 0 (horizontal projectile) a x = 0a y = m/s 2

Step 2: Write down equations that might solve it.

Step 3: Carry out the solution. Time for the jump Therefore, using the modified y equation to solve for the time of the jump.

3.3 Projectile Motion b) What are the magnitude and direction of the final velocity of the Care package? Practice Problem # 2 - A Falling Care Package

Practice Problem # 1 Step 1: Write down what’s Given and your unknowns. (remember to separate your X and Y components using a table) X- ComponentY - Component Dx =Dy = m (negative due to downward direction of fall) v ox = 115 m/sv oy = 0 (horizontal projectile) a y = m/s 2 t = 14.6 s

Step 2: Write down equations that might solve it.

3.3 Projectile Motion Step 3: Carry out the solution. Final vertical velocity of the package

Step 4: Carry out the solution. Magnitude and direction of final velocity

CLASS ASSIGNMENT NOT H/W I will not accept late work

3.3 Projectile Motion Conceptual Example 5 I Shot a Bullet into the Air... Suppose you are driving a convertible with the top down. The car is moving to the right at constant velocity. You point a rifle straight up into the air and fire it. In the absence of air resistance, where would the bullet land – behind you, ahead of you, or in the barrel of the rifle? Explain

Section 3. Cont. Projectiles at an Angle

Projectiles Launched at an angle As shown below, projectiles launched at an angle have an initial vertical and horizontal component of velocity.

Projectiles Launched at an angle

3.3 Projectile Motion Example 6 The Kickoff A placekicker kicks a football at and angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, calculate the following: a)determine the maximum height that the ball attains.

Step 1: Write down what you know and don’t know (and resolve the vector provided) X – ComponentY - Component a x = 0m/s 2 a y = m/s 2  y = ?  t = ?  x = ? =22 m/s =40 o

Step 2: Decide on a plan or equation to use. For the horizontal velocity:  For the vertical velocity:

Step 3: Carry out the plan. yayay vyvy v oy t ?-9.81 m/s m/s

3.3 Projectile Motion Example 6 The Kickoff A placekicker kicks a football at and angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, calculate the following: a)The maximum height that the ball attains = 10 m b)The time between kickoff and landing

Step 1: Write down what you know and don’t know (and resolve the vector provided) X – ComponentY - Component a x = 0m/s 2 a y = m/s 2  x = ?  y = 10.1 m  t = ?   v oy v ox V i = 22 m/s

Step 2: Decide on a plan or equation to use. We can calculate tine for the vertical displacement:

For the vertical displacement (to maximum height): = 2.87 s Step 3: Carry out the plan.

3.3 Projectile Motion Example 6 The Kickoff A placekicker kicks a football at and angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, calculate the following: a)The maximum height that the ball attains = 10 m b)The time between kickoff and landing = 2.86 s c)The range ‘R’ of the projectile?

Step 1: Write down what you know and don’t know (and resolve the vector provided) X – ComponentY - Component a x = 0m/s 2 a y = m/s 2  x = ?  y = 10 m  t = 2.86 s  t = 1.43 s   v oy v ox V i = 22 m/s

Step 2: Decide on a plan or equation to use. For the horizontal displacement:

= 48.5 m Step 3: Carry out the plan.

Practice Problem A baseball is thrown at an angle of 25 o relative to the ground at a speed of 23.0 m/s. If the ball was caught 42.0 m from the thrower, how long was it in the air? How high was the tallest spot in the ball’s path?

Step 1: Write down what you know and don’t know (and resolve the vector provided) X – ComponentY - Component a x = 0m/s 2 a y = m/s 2  x = 42m  y = ?  t = ?   v iy v ix V i = 23 m/s

Step 2: Decide on a plan or equation to use. For the horizontal displacement:  For the vertical displacement:

Step 3: Carry out the plan. v i =23m/s v ix

Step 3: Carry out the plan. The highest point will be half-way through the flight.  t = ½ x 2.0 s = 1.0 s v i =23m/s v iy

Class Assignment In 1991, Doug Danger rode a motorcycle to jump a horizontal distance of 76.5 m. Find the maximum height of the jump if his angle with respect to the ground at the beginning of the jump was 12.0°. 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, what is the magnitude of the flea’s horizontal displacement? How high does the flea jump?