Free Fall Kinematics PDF

Free Fall UNIFORMLY ACCELERATED MOTION  Objectives At the end of the lesson, students should be able to: 1. define free fall; 2. describe the behavior of a free-falling object; 3. characterize the three cases of free fall; 4. derive the kinematic equations for free fall; and 5. solv...

Free Fall UNIFORMLY ACCELERATED MOTION  Objectives At the end of the lesson, students should be able to: 1. define free fall; 2. describe the behavior of a free-falling object; 3. characterize the three cases of free fall; 4. derive the kinematic equations for free fall; and 5. solve problems involving free fall. PREDICT-OBSERVE-EXPLAIN Which of the two objects will reach the ground first? WITH AIR RESISTANCE WITHOUT AIR RESISTANCE  Debunking Common Myth The Myth: When dropped at the same height, heavier objects fall faster than light objects because of their mass.  Debunking Common Myth The Truth: Light objects with greater surface area tend to fall ‘slower’ because of air resistance (air friction).  Debunking Common Myth The Truth: In absence of air resistance, all objects will fall at the same rate regardless of the object’s mass.  Free Fall Free fall is the movement of an object under the influence of gravity only - without other forces acting on it, such as air resistance or friction.  Key Features of a Free Falling Object 1. Acceleration due to Gravity The only force that acts on a free-falling object is the gravitational force. The acceleration is downward and towards the Earth’s center. 𝑚 ⃗ ⃗ 𝑎 = 𝑎 𝑔=− 9.8 2 𝑠  Key Features of a Free Falling Object 2. Time Symmetry (for objects thrown upward) The time required for an object to reach its maximum height is equal to the time for it to return to its starting point. 𝑡 𝑢𝑝𝑤𝑎𝑟𝑑 =𝑡 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑  Key Features of a Free Falling Object 3. Speed Symmetry (for objects thrown upward) At any point of release, the speed of the body during the upward trip is the same as the speed during the downward trip. 𝑣 𝑢𝑝𝑤𝑎𝑟𝑑 =𝑣 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑  Key Features of a Free Falling Object 4. Velocities (for objects thrown upward) Upward velocity is positive and decreasing, while downward velocity is negative but increasing. ( ⃗𝑣↑ , 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒𝑠 ; ⃗𝑣↓ ,𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑠 )  The Kinematic Equations for Free Fall Since free-falling objects have the same acceleration (), the horizontal kinematic equations shall be utilized but with a slight change in the variables. The Kinematic Equations for Free Fall in terms of the acceleration If is not given If is not given If t is not given 𝑣 0 𝑦= √⃗ 𝑎𝑔 ∆ ⃗𝑦 𝑣 𝑓 𝑦 =√ 𝑣 0 𝑦 +2 ⃗ 2 ⃗ 2 𝑣 𝑓𝑦 −2 ⃗ ⃗ ⃗ 𝑎𝑔 ⃗𝑦  Three Cases of Free Fall 1. Object Dropped at a Certain Height 2. Object Thrown Downward 3. Object Thrown Upward  Object Dropped at a Certain Height An object dropped at a certain height has NO INITIAL VELOCITY. ⃗ 0 𝑦 =0 𝑚 / 𝑠 𝑣  I. Object Dropped at a Certain Height Sample Problem: A rock is dropped from the top of a cliff that is 78.4 meters high. How long does it take to reach the ground? Given: Equation: Solution: 1⃗ 2 2 ∆⃗ 𝑦 =⃗ 𝑣0 𝑦 𝑡 + 𝑎𝑔 𝑡 2 ∆ ⃗ 𝑦 ⃗ 𝑎𝑔 𝑡 2 = ⃗ 𝑎𝑔 ⃗ 𝑎𝑔 √ 2∆ ⃗ 𝑦 √ 1⃗ 2 ∆ 𝑦= (2)⃗ 𝑎 (2) 𝑡 =𝑡 2 ( − 78.4 𝑚 ) 2 𝑔 ⃗ 𝑡= Required: 𝑎𝑔 𝑚 − 9.8 2 2 𝑠 2∆ ⃗ 𝑦=⃗ 𝑎𝑔 𝑡 𝑡 =4.0 𝑠 Answer:  Object Thrown Downward An object thrown downward has a NON- ZERO NEGATIVE INITIAL VELOCITY. (where x is any value of the initial velocity, depending on the context of the problem)  II. Object Thrown Downward Sample Problem: You threw a ball downward from a window at a speed 2.0 m/s. How fast will it be moving before it hits the ground from a height 2.5 m? Given: Equation: Solution: 2 2 ⃗ 𝑣 𝑓 𝑦 =⃗ 𝑣 0 𝑦 +2 ⃗ 𝑎𝑔 ∆ ⃗𝑦 𝑣 𝑓 𝑦 =√ ⃗ √( 2 ⃗ 𝑣 0 𝑦 +2 ⃗ 𝑎𝑔 ∆ ⃗𝑦 ⃗ ) ( ) 2 𝑚 𝑚 𝑣 𝑓𝑦 = − 2.0 +2 − 9.8 2 ( − 2.5 𝑚 𝑠 𝑠 Required: 𝑚 𝑣 𝑓 𝑦 =− 7.28 ⃗ 𝑠 Answer:  Object Thrown Upward An object thrown upward has a NON-ZERO POSITIVE INITIAL VELOCITY. (where x is any value of the initial velocity, depending on the context of the problem)  Object Thrown Upward An object thrown upward follows time and speed symmetry. 𝑡 𝑢𝑝𝑤𝑎𝑟𝑑 =𝑡 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑 (magnitude only)  Object Thrown Upward The velocity at the maximum height is zero. ⃗ max h𝑒𝑖𝑔h𝑡 =0 𝑚/ 𝑠 𝑣  Object Thrown Upward Its upward velocity is positive and decreasing, while the downward velocity is negative but increasing. ( ⃗𝑣↑ , 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒𝑠 ; ⃗𝑣↓ ,𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑠 )  Object Thrown Upward (Problem Solving Strategy) Solve for the first half of the trajectory and treat the second half of the trajectory as Case 1 (object dropped at a certain height).  III. Object Thrown Upward Sample Problem: An object is thrown 3.5 m/s upward. What is the maximum height reached by the object and the time it remains in the air before landing back to its original position? Given: @1st half: Equation: Solution: Required: Answer:  III. Object Thrown Upward Sample Problem: An object is thrown 3.5 m/s upward. What is the maximum height reached by the object and the time it remains in air before landing back to its original position? Following time Given: symmetry: @1st half: Solution: Equation: Required: Answer:  Modified True or False The motion of a free-falling object depends on its mass. does not depend  Modified True or False A free-falling object experiences no air resistance, allowing it to accelerate solely due to gravity. TRUE  Modified True or False The velocity of a free-falling object is at its maximum when at the maximum height. 𝑍𝐸𝑅𝑂  Modified True or False The kinematic equations can be applied for free-falling objects because it is an example of uniform motion. Uniformly Accelerated Motion  Modified True or False The upward velocity of a free-falling object is decreasing and increasing when moving downward. TRUE  Free Fall Sample Problem 1: A coin was dropped into a deep well and was heard to hit the water after 3.41 s. Determine the depth of the well and the final velocity of the coin just before it hits the water. Given: Equation: Solution: 1 𝑣 0 𝑦 =0 𝑚/ 𝑠 a. ∆ 𝑦 =𝑣 0 𝑦 𝑡 + 𝑔𝑡 2 2 2 𝑔=− 9.8 𝑚/ 𝑠 2 𝑡 =3.41 𝑠 1 𝑚 ∆ 𝑦 =0+ (− 9.8 2 )(3.41 𝑠) 2 𝑠 Required: ∆ 𝑦=−56.98m a. depth of the well () b. final velocity () Answer:  Free Fall Sample Problem 1: A coin was dropped into a deep well and was heard to hit the water after 3.41 s. Determine the depth of the well and the final velocity of the coin just before it hits the water. Given: Equation: Solution: 𝑣 0 𝑦 =0 𝑚/ 𝑠 2 b. b. 𝑔=− 9.8 𝑚/ 𝑠 𝑡 =3.41 𝑠 2 𝑣 𝑓𝑦 =0 +− 9.8 𝑚/ 𝑠 ( 3.41 𝑠) 𝑣 𝑓𝑦 =− 33.42 𝑚/ 𝑠 Required: a. depth of the well () b. final velocity () Answer:  Free Fall Sample Problem 2: In 1984, Michael Jordan soared to a height of 48 inches. It was one of the top highest vertical leaps. Calculate his takeoff speed and hangtime. Given: Equation: Solution: 0.0254 𝑚 𝑣 𝑜𝑦 =√ 𝑣 𝑓𝑦 − 2𝑔 ∆ 𝑦 2 2 𝑣 𝑓𝑦 − 𝑣 𝑜𝑦 2 𝑦 =48 𝑖𝑛 𝑥 =1.22 𝑚 ∆ 𝑦= 21 𝑖𝑛 2𝑔 𝑔=9.8𝑚 /𝑠 2 2 𝑔 ∆ 𝑦 =𝑣 𝑓𝑦 − 𝑣 0 𝑦 2 𝑣 =√ 0 − 2(−9.8𝑚 /𝑠 )(1.22𝑚) 𝑜𝑦 2 2 𝑣 𝑦 𝑎𝑡 𝑡h𝑒max h𝑒𝑖𝑔h𝑡=0 - 𝑣 𝑜𝑦 =4.89 𝑚 / 𝑠 Required: 𝑣 𝑜𝑦 =√ 𝑣 𝑓𝑦 − 2𝑔 ∆ 𝑦 2 a. takeoff speed () a. hangtime (T) Answer:  Free Fall Sample Problem 2: In 1984, Michael Jordan soared to a height of 48 inches. It was one of the top highest vertical leaps. Calculate his takeoff speed and hangtime. Given: Equation: Solution: 0.0254 𝑚 𝑇 =2 𝑡 𝑦 =48 𝑖𝑛 𝑥 =1.22 𝑚 2 1 𝑖𝑛 𝑔=9.8𝑚 /𝑠 𝑇 =2 (0.5 𝑠) 𝑣 𝑦 𝑎𝑡 𝑡h𝑒max h𝑒𝑖𝑔h𝑡=0 𝑇 =2 𝑡 Required: a. takeoff speed () a. hangtime (T) Answer: T= 1s  Free Fall Sample Problem 3: A stone is thrown downward from a 5-m high window. It reached the ground after 0.5 s. What is the initial and final velocities of the stone? Given: Equation: Solution: 1 ∆𝑦 1 ∆ 𝑦 =𝑣 𝑜𝑦 𝑡 + 𝑔𝑡2 𝑣 𝑜𝑦 = − 𝑔𝑡 2 𝑡 2 1 2 ∆ 𝑦 − 𝑔 𝑡 =𝑣𝑜𝑦 𝑡 ) 2 1 𝑣 𝑜𝑦 = −5𝑚 0.5 𝑠 − 1 2 ( 𝑚 𝑠 ) − 9.8 2 ( 0.5 𝑠 ) 𝑡 𝑚 Required: 𝑣 𝑜𝑦 =− 7.55 ∆𝑦 1 𝑠 − 𝑔 𝑡 =𝑣 𝑜𝑦 𝑣 𝑜𝑦 =? 𝑡 2 Answer:  Free Fall Sample Problem 3: A stone is thrown downward from a 5-m high window. It reached the ground after 0.5 s. What is the initial and final velocities of the stone? Equation: Solution: Given: 𝑣 𝑓𝑦 = 𝑣 𝑜𝑦 + 𝑔𝑡 = = Required: 𝑣𝑓 𝑦 =? Answer: