Free fall is the unimpeded motion of an object in the absence of air friction (resistance) where only gravitational force influences the object.
Gravitational acceleration:
Gravitational acceleration (g ) is the constant acceleration of a free falling object due to gravity.
All objects experience the same gravitational acceleration (if we ignore the effects of air resistance). Hence, if there were no air a feather would fall at the same speed as a stone. See http://www.youtube.com/watch?v=5C5_dOEyAfk (where O is "oh", not zero).
It is always directed downwards.
On earth it is gearth = 9,8 m·s–2 and on the moon:gmoon = 1,6 m·s–2
Formulas All the following formulas are useful when you calculate projectile motion:
where: v i is initial velocity (m·s−1) v f is final velocity (m·s−1) Δ y is displacement (m) a is acceleration (m·s−2) Δt is time (s)
Tips for calculations:
Ignore air resistance for all calculations in Grade 12 unless the question states that there is air resistance.
Free falling objects experience a constant downward acceleration equal to the gravitational acceleration, (g = 9,8 m·s–2).
Choose a direction (downwards or upwards) as positive and keep this unchanged throughout the problem.
Indicate the direction you have chosen as positive clearly at the start of your answers.
REMEMBER: Gravitational acceleration “g ” is constant and always directed downwards evenwhen the object is moving upwards. Therefore:
If upwards direction is chosen as positive gravitational acceleration (g) will be negative.
If downwards direction is chosen as positive gravitational acceleration (g ) will be positive.
projectile: an object (e.g. stone, ball or bullet) that travels through the air while gravity is the only force acting on it. unimpeded: without being opposed or obstructed or disturbed.
When a projectile is dropped (from rest) from a certain height, then:
Initial velocity vi = 0 m·s–1
The velocity increases as the object falls downwards
The velocity is a maximum (vf) as the object hits the ground.
Graphs and Projectile Motion: Type 1 – Dropping a projectile Downwards as positive Upwards as positive The distance (y) starts increasing slowly and then more quickly.
The velocity (v) increases at the same rate from start to finish.
The accelertion is constant at 9,8 m/s2 whether it is in a positive direction or not.Only the sign changes. NOTE: If the object is thrown downwards (not dropped) from a certain height, then initial velocity is not zero. vi ≠ 0 v i = the velocity at which the object is thrown.
NOTE: Note that the ground is taken as the zero reference for these graphs.
Activity 1 A ball of mass 200 g is dropped from the roof of a 100 m high building. Ignore air resistance and calculate:
the velocity of the ball when it hits the ground. (4)
how long the ball is in the air before it hits the ground. (5) [9]
Solutions (Calculations for ‘down positive’ and for ‘up positive’ are provided. You only need to do one way!) Let direction of motion down be positive vi = 0 m·s−1 vf = ? (a) Δy = + 100 m a = + 9,8 m·s−2 Δt = ? (b)
vf = vi + a Δt 44,27 = 0 + (9,8) Δt Δt = 44,27 = 4,52 s 9.8 ∴ the ball is in the air for 4,52 s (5) [9]
If we let direction of motion up to be positive the solution is the same, only the sign changes. This example shows that projectiles can have their motion described by a single set of equations for both upward and downward motion. It is not necessary to set motion in two directions for the same question. However it is important that you are able to solve problems using both approaches i.e. downwards - positive OR upwards - positive.
The mass of a falling object is irrelevant during free fall. Ignore the given value! The value of the mass is only relevant if you’re asked to calculate the momentum with which it strikes the ground. Remember the feather and hammer.
Thinking about Physics A gun fires a bullet up into the air and the bullet leaves the gun at mach 1 (the speed of sound). Suppose the bullet flies up to a certain height and then falls down to earth. Will it hit the ground at mach 1, or will it reach a certain limiting velocity and not achieve mach 1 again? Answer: It will hit the ground at approximately mach 1 (ignoring air resistance). The reason is that as it flies up, it decelerates (gets gradually slower), until it reaches 0 m/s at the peak of its travel. It then has the same Ep (potential energy) as it did when it left the gun with the original kinetic energy (Ek). Thus, as it falls back to earth, the Ep is converted to Ek again. Since energy is conserved, the amount of energy with which it strikes the ground, must be the same as that energy that it had when it left the gun, namely, enough energy to reach mach 1. The deceleration of the bullet from mach 1 to 0 m/s is entirely due to gravity and air resistance; hence, when it falls back, its acceleration will be entirely due to gravity. NB: These three graphs below are of the SAME EVENT: an object thrown upwards, then falling back down.
Displacement - time graph
The displacement (y) increases in the positive direction until it reaches a maximum, and then it decreases. It becomes negative if it moves below the starting point. NB: This graph tracks the position of an object thrown up, then falling, in time. It DOES NOT mean that the object is thrown in an arc.
Velocity - time graph
The velocity (v) decreases until it reaches zero at the maximum height. Then the magnitude of the velocity increases, but in a negative direction, hence it goes below the t axis. NB: This is NOT a rock rolling down a hill. This is an object being thrown or shot up, reaching a maximum height, then falling back down again.
Acceleration - time graph
The acceleration is constant at 9,8 m/s2 throughout, since we chose upwards as positive. NB: This is NOT an object travelling in a straight line; it is an object experiencing uniform negative acceleration (deceleration) due to gravity.
For the three graphs illustrated above
Upwards is positive. If downwards is taken as positive the graphs will be inversed (upside-down).
The original position is taken as the reference point.
Initial velocity (vi) at A. The object leaves the starting point in an upwards direction.
v l is negative, as it is moving upwards.
The magnitude of the velocity decreases as the object rises
The velocity is zero when the object reaches the highest point at B.
vf (up) at B = 0 = v i (down) at B
The velocity increases as the object falls down towards the ground.
The initial velocity up is equal in magnitude to the final velocity down.
vi (up) at A= v f (down) at C Note: vi (up) at A = vf (down) at C because level A is the same as level C If upwards is chosen as positive g = –9,8 m·s−2 ALWAYS.
If downwards is chosen as positive g = +9,8 m·s−2 ALWAYS.
The time taken to rise from A to B = time taken to return from B to the original position C t (up) AB = t (down) BC
The total time taken through AB to C = time to rise from A to B + time to return from B to original position C.
The object’s displacement is zero (as it returns to its original position).
The velocity (vf) down is a maximum as the object hits the ground at C.
The acceleration of the object is constant at g = 9,8 m·s−2 downwards throughout the motion.
Activity 2 A ball is thrown vertically upwards at 4 m·s−1 and returns to the thrower’s hand. Let the direction of motion downwards be positive. Calculate:
The maximum height reached by the ball. (4)
The time taken for the ball to reach the highest point in its trajectory. (5)
The total time that the ball is in the air. (3)
The ball’s total displacement during the motion. (1) [13]
Solutions Let the direction of motion down be positive vi = −4 m·s−1 vf = 0 m·s−1 Δy = ? (a) m a = + 9,8 m·s−2 Δt = ? (b)
v2f = v2i + 2a · Δy 0 = (–4)2 + (2)(9,8) yΔ 19,6 Δy = –16 Δy = –0,82 m ∴ the ball reaches a height of 0,82 m above the starting level (4)
vf =vi + a Δt 0 = (–4) + (9,8)Δt Δt = 4 = 0,41 s 9.8 ∴ the ball takes 0,41 s to reach the highest point in its trajectory (5)
Time upwards = time downwards ∴ total time in the air is (2)(0,41) = 0,82 s (3)
Total displacement = Δy = 0m (1) Displacement is measured in a straight line from the initial position (the thrower’s line from the original to the final position (the thrower’s hand is the initial and final position). [13]
At the highest point of the trajectory the velocity is 0 m·s–1 while the acceleration is still 9,8 m·s–2 downwards.
is velocity as object leaves the starting point in an upwards direction. It is positive.
The magnitude of the velocity decreases as the object rises.
The velocity is zero when the object reaches the highest point at B i.e. vf (up) at B = 0 = v i (down) at B.
The magnitude of the velocity increases as the object falls downwards.
The magnitude of the initial velocity upwards at A is equal to magnitude of velocity down at the starting level, C.
v i (up) at A = −v f (down at starting level) at C. If downwards is chosen as positive g = +9,8 m·s−2 ALWAYS.
If upwards is chosen as positive g = –9,8 m·s−2 ALWAYS.
The time taken to rise from A to B = time taken to return from B to the original position C.
The total time taken to move though ABC to D = time to rise from A to B + time to return from B to original position C + time from C to the ground D.
The object’s displacement is downwards (as it passes the starting point and falls further downwards).
The object’s displacement (A to D) is equal in magnitude to the height from which it was released The final velocity (vf) is a maximum as the object hits the ground at D.
The acceleration of the object is constant, g = 9,8 m·s−2 downwards throughout the motion.
Activity 3 Lerato throws a stone vertically into the air from the top of a cliff. The stone strikes the ground below after 3 s. The velocity vs. time graph below shows the motion of the stone. Ignore the effect of air resistance.
Hint: Use a ruler to draw the axes and any straight lines! Drawing a sketch graph
The graph does not have to be to scale, but it must have the correct shape
The graph must show the physical quantity in words followed by the abbreviated unit (in brackets) e.g. velocity (m·s−1)
The graph must include the values asked for in the question.
How long does the stone take to fall from the height of the cliff to the ground below? (2)
What is the maximum height that the stone reaches above the groud? (Hint: calculate the height the stone reaches above the cliff, then calculate the height of the cliff, and add these two numbers). (4)
Draw a graph of position versus time. Use upwards as negative. (6) [12]
Solutions
3 - 2,043 = 0,96 s (2)
Option 1 Δy = vi Δt + ½ a Δt2 = (10)(3) + ½ (–9,8)(3)2 = 14,1 m Δy = 14,1 m below the starting point vf2 = vi2 + 2aΔy 0 = 100 + 2(–9,8) Δy Δy = 5,1 m Maximum height above the ground = 5,1 + 14,1 = 19,2m Option 2 Δy = vi Δt + ½ a Δt2 = 0 3 + ½ (–9,8)(3 – 1,02)2 = –19,21 m Δy = 19,21 m (maximum height above the ground) Option 3 vf2 = vi 2 + 2aΔy (–19,4)2 = 0 + 2 (–9,8) Δy Δy = 19,2 m (maximum height above the ground) (4)
Marks for: correct shape; graph starts at zero; maximum height shown as –5,1 m ; times indicated correctly; graph ends at 3 s . (6) [12]
When a ball falls freely through the air, gravity is the only force that acts on it. The resultant force on the ball is downwards and it accelerates in the direction of the resultant force (Newton’s Second Law of Motion).
Consider a ball that bounces up from the ground (at A). Let direction of motion upwards be positive.
The ball rises to a maximum height while slowing down (gravity accelerates it downwards at 9,8 m·s−2)
At the highest point in its path (at B) its velocity is 0 m·s−1
It still accelerates downwards at 9,8 m·s−2 due to gravity.
It falls to the ground and its velocity increases until it strikes the ground (at C).
From C to D the ball is in contact with the ground.
The ground exerts an upward force on it which is greater than the force of gravity.
The resultant force on the ball is therefore upwards and the ball accelerates upwards, in other words it slows down and stops.
While still in contact with the ground, it starts to move upwards and (at D) leaves the ground.
The collisions with the ground are inelastic and some of the ball’s kinetic energy is transformed into e.g. sound and heat every time it strikes the ground.
So the velocity as the ball leaves the ground (v i upwards) is less than the velocity at which it hits the ground.
After each bounce, the height reached by the ball is less than during the previous bounce.
The mass of the ball is irrelevant. The ball is in free fall – this means that all effects of air resistance may be ignored.
Activity 4 A hot-air balloon is rising upwards at a constant velocity of 5 ms–1. When the balloon is 60 m above the ground, a boy drops a ball from it and the ball falls freely. Assume that the balloon continues to move upwards at the same constant velocity. When the ball hits the ground, it bounces vertically upwards to a height of 8 m above the ground. It falls back to the ground and bounces again to reach a height of 5 m. Take upwards as positive.
Describe the motion of the ball from the moment it is dropped until it hits the ground. (8)
Why does the ball not reach the same height during the second bounce as during the first? (7)
What is the magnitude and direction of the ball’s velocity at the moment when it is dropped? (1)
Calculate maximum height reached by the ball. (5)
Where is the ball after 3 seconds? (6)
How far apart will the ball and the balloon be after 3 seconds? (7)
Calculate the time taken for the ball to reach the ground. (7)
Calculate the time the ball takes to reach the height of 8 m above the ground after its first bounce. (10)
Calculate the velocity at which the ball hits the ground the after the first bounce. (4)
Draw a sketch graph of velocity vs. time for the ball from the moment it is dropped until it reaches the height of 5 m after its first bounce. (5)
Draw a sketch graph of position vs. time for the ball for the same time as in (10). Use the position of the ball when it is dropped as the point of reference. (5)
Draw a sketch graph of acceleration vs time for ball for the same time as in (10). (4) [69]
Solutions
Initially the ball and hot-air balloon will both move upwards at a constant velocity. When the ball is dropped it continues to move upwards but decelerates constantly (at 9,8 m·s−2) due to the gravitational attraction force of the earth and slows down until it reaches the highest point in its trajectory (path). It stops momentarily ( v = 0) and then starts to accelerate downwards constantly (at 9,8 m·s−2). Its speed increases until it hits the ground at a maximum velocity. (8)
The collision between the ball and the ground is inelastic. Some of the ball’s kinetic energy is converted into heat and sound energy and the ball is deformed during the collision. The upward force of the ground on 3 the ball causes it to bounce upwards but the kinetic energy is less than before the collision, so the velocity 3 at which the ball leaves the ground is less than the velocity at which it hit the ground and the height reached is lower 3 than the previous bounce. (7) The point of reference for a position-time graph is placed on the time axis, where y = 0 m.
5 ms–1 upwards (1)
vi = +5 ms–1 vi = 0 ms–1 a = –9,8 ms–2 vf2 = vi2 + 2 a ·Δy 02 = 52 + 2(–9,8)Δy –25 = –19,6Δy Δ y = -25 = 1,28 m -19.6 ∴ the ball will reach a maximum height of (60 + 1,28) = 61,28 m above the ground. (5)
vi =+5 ms–1 Δt = 3 sa = –9,8 ms–2 Δy = vi Δt + ½ at2 ∴ Δy = (5)(3) + ½ (–9,8)(3)2 ∴ Δy = – 29,1 m ∴ the ball is 29,1 m below the point from where it was released, or (60 – 29,1) = 30,9 m above the ground. (6)
The hot-air balloon moved upwards at a constant velocity. Δy = vi Δt + ½ a Δt2 Δy = (5)(3) + 0 Δy = 15 m ∴ After 3s the hot-air balloon will be 15 m above the starting point. We know from Question 4 that the ball will be 29,1 m below the starting point after 3s. ∴ after 3s the hot-air balloon and the ball will be (15 + 29,1) = 44,1 m apart. (7)
NOTE: Always calculate the velocity at which the ball hits the ground first. vi = + 5 ms–1 Δy = – 60 ma = –9,8 ms–2 vf2 = vi2 + 2a ·Δy vf2 = (5)2 + 2(–9,8)(–60) ∴ vf = 34,66 ms–1 downwards then vf = vi + a ·Δt –34,66 = 5 + (–9,8)Δt ∴ Δt = 4,05 s (7)
NOTE: The up and down displacement of the ball from the first to the second contact with the ground, is the same in magnitude (size) and tup = tdown. Consider the downward motion as negative as in the previous calculations. For the downward part of the bounce:
vi = 0 ms–1 Δy = – 8 m a = – 9,8 ms–2 Δy = vi Δt + ½ a Δt2 – 8 = 0 + ½ (–9,8)(Δt2) Δt = 1,28 s but tup = tdown ∴ the ball takes 1,28 s to reach a height of 8 m. (10)
vf = vi + a Δt vf = 0 + (–9,8)(1,28) vf = –12,54 ms–1 ball hits the ground at 12,54 ms–1 downwards after the first bounce (4)
(5)
(5)
(4) [69]
Activity 5
A 30 kg iron sphere and a 10 kg aluminium sphere with the same diameter fall freely from the roof of a tall building. Ignore the effects of friction. When the spheres are 5 m above the ground, they have the same ...
momentum.
Acceleration
kinetic energy
potential energy (2)
An object is thrown vertically into the air at 12 m·s–1 in the absence of air friction. When the object is at the highest point, the velocity of the object in m·s–1 is?
0
9,8 downwards
9,8 upwards
12 (2)
An object is projected vertically upwards and then falls back to the ground level. The acceleration of the object is ….
Directed upwards throughout its movement.
Zero at the greatest height.
Directed downwards throughout its movement.
Directed upwards and then downwards. (2) [6]
Solutions 1. B (2) 2. A (2) 3. C (2) [6]
Activity 6 A ball of mass 0,15 kg is thrown vertically downwards from the top of a building to a concrete floor below. The ball bounces off the floor. The velocity versus time graph below shows the motion of the ball. Ignore the effects of air friction. TAKE DOWNWARD MOTION AS POSITIVE.
From the graph, write down the magnitude of the velocity at which the ball bounces off the floor. (1)
Is the collision of the ball with the floor ELASTIC or INELASTIC? refer to the data on the graph to explain the answer. (2)
Calculate the:
Height from which the ball is thrown (3)
Size of the displacement of the ball from the moment it is thrown until time t (5) (in an exam, you might see the word “magnitude” — this means “size”).
Sketch a position versus time graph for the motion of the ball from the moment it is thrown until it reaches its maximum height after the bounce. USE THE FLOOR AS THE ZERO POSITION. Indicate the following on the graph:
The height from which the ball is thrown
Time t (4) [15]
Solutions
15 m·s–1 (1)
Inelastic The speed/velocity at which the ball leaves the floor is less / different than that at which it strikes the floor OR The speed/ velocity of the ball changes during the collision. Therefore the kinetic energy changes/is not conserved. (2)
Displacement from floor to maximum height vf2 = vi2 + 2aΔy (0)2 = (–15)2 + 2(9,8)Δy Δy = –11,48 m Total displacement = –11,48 + 15,3 = 3,82 m or 3,83 m (5)
Marking criteria for graph:
Correct shape as shown for first part.
√
Correct shape as shown for the second part up to t / 2,55 s.
√
Graph starts at –15,3 m at t = 0 s.
√
Maximum height after bounce at time t / 2,55 s.
√
Maximum height after bounce less than 15,3 m.
√
(4) [15]
Last modified on Thursday, 23 September 2021 06:11