Vertical Projectile Motion

The following terms are important in vertical projectile motion: • Gravitational field: This is the space around an object of mass (e.g. the earth) in which another mass experiences a gravitational force.  g = 9,8 m·s-2 downwards for the full motion  v downwards increases final velocity v f is downwards    v f = - vi (v f  0) • Time symmetry: The time interval from a certain height to the turning point = time interval from the turning point to the same height. • Acceleration due to gravity (g): The acceleration of a falling object as a result of the attractive gravitational force of the earth, in the absence of other forces such as air resistance. • At the same time interval t on either side of the turning point, the instantaneous velocity is the same, but in the opposite direction.   • The displacement ( y) is the linear distance from an initial position (yi ) to a  final position (y f ) for the specific part of the motion that is being worked with. • Free fall: The unhindered movement of an object in the gravitational field of the earth, when only the gravitational force is acting on it. • Projectile: An object that is moving through the air, experiencing free fall, after being dropped, thrown or shot. (In the air, air resistance is small. If air resistance is ignored, gravitational force is the only force acting on the object.)  • The displacement ( y) can be upwards or downwards with respect to the  initial position. Use the correct sign for y according to how the direction was chosen, i.e. +/- or vice versa. When returned to the starting point,  y = y f - yi = 0. Representation of three different types of projectile motion Description of projectile motion Describe projectile motion in terms of words, equations and graphs: MECHANICS  Vertical upward  Vertical upward  Vertical downward projection from a projection from a projection from a starting starting point; object starting point above point above the ground. turns around and returns the ground; object turns, The object can: to starting point. moves downwards, past the starting point,  be dropped (v i = 0) to the ground. OR  be thrown downwards at a velocity v i > 0. • In words: Give a description in words of the object's position, initial and final velocities, acceleration, displacement and the duration of time of the motion. • Equations of motion: A projectile motion upwards and downwards can be described by a single set of equations, namely one of the following equations of motion: y vi g 0 Copyright © The Answer Series: Photocopying of this material is illegal 19 Horizontal motion vf = vi + a t   2  vf = vi2 + 2a x Remember: The value of a variable can differ for different parts of the motion, e.g. velocity at turning point is the v f for  and v i for  motion. Vertical motion vf = vi + a t   2  vf = vi2 + 2a y   g = 9,8 m·s-2 is known already; therefore we only need the values of any two other variables in order to calculate the missing value in an equation.   1 x = vi t + a t 2   1 y = vi t + a t 2  Choose the correct equation.     v + vf  x =  i  t  2      v + vf  y =  i  t  2   Remember the signs of vi, v f and 2  vi  vf  a  g where:  x ; 2    y.   The sign for g will be the same as that of the downward motion. • Do substitution in the applicable equation. = initial velocity (m·s-1) • Do the calculation and give the answer with the correct unit and direction. = final velocity (m·s-1) = acceleration (m·s-2) Remember:   v i = 0 m·s-1 if the object falls from rest OR   v i = the velocity of another moving object off which the object falls = gravitational acceleration = 9,8 m·s-2 t = time (s)   y = displacement (m) ( x for horizontal, 1 NOTES Equations of motion  y for vertical)  the velocity is 0 m·s-1 at the turning point  the velocity at which the object falls increases during downward  free fall. The velocity at which it hits the ground (v f ) g 0, but it has Application: Problem-solving steps a maximum value. • Read the problem carefully and draw a sketch with all the necessary information. • Determine with which part of the motion a specific question is concerned, e.g. with the full motion or only the upward motion up to the turning point, or with a certain time interval, etc. MECHANICS • Choose the direction of motion if working with both upward and downward motion in one calculation, e.g. if upwards is chosen as + , downwards is taken as -.  Every equation of motion contains four of the five possible variables,     i.e. vi, v f , g, t, y.  Make a list of all the variables and fill in the values provided. 20 Copyright © The Answer Series: Photocopying of this material is illegal Graphs of motion for vertical motion Projectile motion up- and downward, BACK TO STARTING POSITION 0 The ground is taken as reference point (zero) Time (s)  v Gradient = t  - -v 0 + - -v Time (s) +  v 0 1 bh 2 + -v v + 0 -v - Time (s) +  v  Gradient = t  y = area under graph Acceleration-time (for all three cases) Time (s) 0 - 9,8 Copyright © The Answer Series: Photocopying of this material is illegal Time (s) + -  = a = -9,8 m·s-2 / +9,8 m·s-2   y = area under graph = sum of the triangles between graph and horizontal axis = 0 (the triangles are the same size, but in opposite directions; therefore they cancel one another out) = area of Acceleration (m·s-2) Time (s) v   y = area under graph MECHANICS + Velocity-time = a = -9,8 m·s-2 / +9,8 m·s-2 = -9,8 m·s-2 / +9,8 m·s-2 Time (s) The starting position was taken as reference point (zero). +  v  Gradient = t = a = + v 0 0 If the turning point or end position is taken as the reference point, the entire graph moves up or down to that position. Velocity (m·s-1) + 0 Time (s) Velocity-time Velocity (m·s-1) Velocity (m·s-1) Velocity (m·s-1) Time (s) 0 + The starting position was taken as reference point (zero). Velocity-time 0 + Time (s) NB: + 0 + Position (m) Time (s) Time (s) Position (m) 0 + Velocity (m·s-1) Time (s) + Position-time 21 = sum of the two triangles between the graph and horizontal axis (the total displacement has the same sign as the larger triangle) Acceleration (m·s-2) 0 Position-time Position (m) + Position (m) Position (m) Position-time Projectile motion up- and downward, y PAST STARTING POSITION Velocity (m·s-1) Vertical downward projectile motion Position (m) NOTES 1 9,8 0 Time (s) + B Position (m) Position-time graph (for uniform acceleration) B  A decreasing gradient shows that the velocity is busy decreasing, e.g. during upward motion.  An increasing gradient shows that the velocity is busy increasing, e.g. during downward motion. Position (m) # Ground as reference point o+ B Time (s) A C Time (s) 0 - + # D D Time (s) 0 C A Note: For position-time graphs:  = position above reference point (indication of direction) o = position below reference point 0 B C • The gradient of the graph gives the acceleration, i.e. 9,8 m·s-2 downwards. The sign of the gradient therefore indicates the sign of the downward motion. + Starting position as reference point Position (m) Position (m) + Time (s) 0 o+ # • The velocity-time graph for an object accelerating uniformly is a straight line. Time (s)  Object is projected upwards and returns to the initial position B + # C o+ # + Time (s) 0 - A vi 0 Time (s)  g = -9,8 m·s-2 + - 0 vi Time (s)  g = +9,8 m·s-2 + • The initial velocity is read on the vertical axis opposite t = 0 (y-intercept) C • The point of intersection with the time axis is the turning point in the air where the velocity = 0 m·s-1. B Starting position (ground) as reference point 22 Copyright © The Answer Series: Photocopying of this material is illegal MECHANICS C 0 A Velocity-time graph (for uniform acceleration) v (m·s-1) B # C + Position (m) 0 Position (m) Time (s) Position (m) Position (m) 0 C B (a position above the ground is the reference point)  Object falls downwards to/from the reference point Time (s) A B • The gradient of the tangent to the graph at any point gives the instantaneous velocity of the object at that moment: C 0 1 (ground as reference point) Time (s) • If the graph moves further away from the reference point (zero position), the displacement increases, and if it moves closer to the reference point, the displacement decreases. B Time (s) 0 o+ # Time (s) D Time (s) • The zero position is the reference point relative to which the object moves, e.g. the ground, the top of a building, etc. # + D Position (m) 0 0 + # C v (m·s-1) Position (m) Position (m) The position-time graph for an upward and downward projectile motion has a parabolic shape. A Position (m)  Object is projected upwards and falls past the initial position NOTES SINGLE PROJECTILE MOTION THE BOUNCING BALL  • If the graph starts on the time axis, e.g. v = 0, it falls downwards from rest. The motion is only in one direction if the object falls or is thrown downwards, e.g.: Ball is dropped from rest at a certain height and bounces on the ground a few times. *  Position-time graph: i)  + 0 + * vf Time (s) vi v (m·s-1) v (m·s-1) Object thrown downwards 0 * + Time (s) t = time ball makes contact with ground 0 vi 0 Time (s) vf turning point Time (s) downward motion • The area between the graph and the horizontal axis indicates the displacement of the object at any given moment. after 2 s: after 4 s: 1 bh 2 1 = (2)v i 2 y = 0 i)  + downward motion upward motion vi 0 2 4 vi 1 y3 0 Time (s) t x2 x1 x3 0 vf 1 x1 x2 x3 Time (s) t y1 y2 y3 • It is in contact with the ground for a very short time interval t. • It bounces back and leaves the ground in the opposite direction at points y1, y2, and y3. = 0m MECHANICS y2 • The ball hits the ground at x 1 , x 2 , x 3 (bounces three times). 1 1 b1h1 + b2 h 2 2 2 1 1 = (2)v i + (2)(- v i ) (v f = -v i ) 2 2 y =  y =0m • The turning point in the air for each bounce is where the graph intercepts the time axis (v = 0). * Note: For velocity-time graphs:  = upward motion (indication of direction) o = downward motion Copyright © The Answer Series: Photocopying of this material is illegal y1 vf 1 - vi 1 Time (s) vf   after 4 s: time symmetry â v f = -v i and ii)  + Time (s) t vi  Velocity-time graph: * + Velocity (m·s-1) t turning point v (m·s-1) 0 * v (m·s-1) v (m·s-1) vi + Time (s) 0 Each bounce is a separate projectile motion of an object, which is projected upwards and returns to the starting position. • If the graph intersects the time axis, the object changes direction at time t. The object first moves upwards (velocity gradually decreases to zero at the turning point) and afterwards moves downwards with increasing velocity. upward motion ii)  + Time (s) Position (m) 0 + Velocity (m·s-1) Time (s) Position (m) * + v (m·s-1) Object falls from rest (v i = 0 m·s-1) v (m·s-1) NOTES 1 • The lines are parallel. The gradient of the lines indicates the acceleration of the ball, i.e. gravitational acceleration (gradient = -9,8 m·s-2 for + and +9,8 m·s-2 for +). 23 Time (s) 0 - 9,8 9,8 Time (s) 0 The ball experiences a constant gravitational acceleration of 9,8 m·s-2 . The sharp points on the graph are the large upward accelerations (due to net force ), when the ball is in touch with the ground for a short time interval. 1 NOTES Remember: A ball experiences an impulse and undergoes a change in momentum ( p) when it hits the ground.     p  Impulse = Fnet t = p  â Fnet =  t   The smaller t, the larger Fnet for the same p.      p = m(v f - v i ) (where v f = velocity with which the ball leaves the  ground /bounces back and v i = velocity with which the ball hits the ground) ii)  +  Definitions  Free fall  Gravitational acceleration  (g = 9,8 m·s-2 downward) PROJECTILE MOTION  Projectile Please note • When a ball moves through the air (free-fall), the only force acting  on it is gravitational force (Fg ).  Downward Fg  Up- and downward; • The ball is only a projectile as it moves through the air, not when it touches the ground. back to starting position  • When the ball hits the ground, the ground exerts an upward force (FN)  Up- and downward; on the ball for a short time interval ( t), there is a resultant/net force,  Fnet > 0, on the ball.  FN  Fg    Bouncing ball  dropped/thrown vertically downward or thrown upward; bounces on the ground several times  each bounce is: - an inelastic collision - a separate projectile motion past starting position  Describe (in words) Fnet = FN + F g (vector addition) upwards  constant acceleration  i.e. g = 9,8 m·s-2 downward  for  motion: • The ball has a maximum velocity and kinetic energy Ek just before it hits the ground. velocity decreases to 0 m·s-1 at turning point  for  motion: velocity increases up to a maximum just before it hits the ground • During each bounce, some of the Ek is converted into other forms of energy. • The ball will therefore bounce back with a smaller magnitude of velocity than that with which it hit the ground. It will therefore bounce back to a lower height than with the previous bounce. Calculate Choose direction, e.g. + and  Use equations of motion and calculate, for example: - maximum height reached - height above ground from which object is projected - time taken to maximum height or until ground is reached - velocity at a certain time, etc. Graphs  position/displacement-time ( y-t)  velocity-time (v-t)  acceleration-time (a-t) • Each bounce is a separate projectile motion. Work with each bounce/projectile motion separately to solve problems. 24 Calculations: (v-t graph)   v  gradient (m) = a = t  area under graph = displacement Copyright © The Answer Series: Photocopying of this material is illegal MECHANICS Acceleration (m·s-2) i)  + Acceleration (m·s-2)  Acceleration-time graph: NOTES 1 Question 2 EXAMPLES A cannonball is fired at a velocity of 25 m·s-1. Calculate: Question 1 A 20 kg concrete block falls off a scaffolding on a building site, hits a pile of sand and comes to a standstill within 10 ms. The average net force of the sand and other forces on the concrete block is 10 500 N upwards. Take upwards +. 1.1 Calculate the impulse of the net force on the concrete block. 1.2 Calculate the velocity at which the block hits the sand before it comes to a standstill. 1.3 Calculate the height of the scaffolding. 1.4 Calculate the total time from when the block starts to fall until it comes to a standstill. 2.1 the maximum height that it reaches. 2.2 the time elapsed until it returns to the same horizontal level as the top end of the cannon barrel, from where it was fired. 2.3 the height above the ground at which the cannonball has an upwards velocity of 10 m·s-1 (assume the top of the barrel is 1 m above the ground). ANSWER 2.1 vf2 = vi2 + 2g y â 0 = 252 + 2(-9,8) y â 19,6 y = 625 â y = 31,89 m ANSWER 1.1 impulse = Fnet t impulse = â t = = -20vi â vi = 105 -20 = -5,25 m·s-1 2.3 During contact with ground: p = 105 kg·m·s-1 upwards vf = 0 m·s-1 vi = ? t = 0,01 s vf2 = vi2 + 2a y (-5,25) 2 = 0 + 2(-9,8)( y) 19,6 y = - 27,5625 y = - 1,41 m = 1,41 m downwards MECHANICS 1.4 â 19,6 y = 625 - 100 vi = 25 m·s-1 upwards vf = 10 m·s-1 upwards y = ? = 525 â y = 26,79 m = 26,79 + 1 = 27,79 m above the ground For motion through the air: v i = 0 m·s-1 vf = 5,25 m·s-1 downwards = -5,25 m·s-1 g = -9,8 m·s-2 y = ? t = ? Question 3 A hot-air balloon travels upwards at a constant velocity. At a height of 80 m above the ground, someone drops his cell phone from the balloon. It hits the ground after 5 seconds. Ignore the effect of air resistance. vf = vi + g t 3.1 Calculate the velocity at which the balloon travels upwards. 3.2 Calculate the maximum height above the ground reached by the cell phone. 3.3 How long after the cell phone has been dropped, does it pass the point from where it started to fall? 3.4 How long after the cell phone has been dropped, does the distance between the balloon and the cell phone equal 20 m? â -5,25 = 0 + (-9,8) t â t = 0,54 s â total time until it comes to a standstill = 0,54 + 0,01 = 0,55 s Copyright © The Answer Series: Photocopying of this material is illegal vf2 = vi2 + 2g y â 102 = 252 + 2(-9,8) y = 5,25 m·s-1 downwards 1.3 25 9,8 = 2,55 s â total time = (2,55)(2) = 5,10 s p = m(vf - vi) â 105 = (20)(0 - vi) vf = vi + g t â 0 = 25 + (-9,8) t m = 20 kg t = 10 ms = 0,010 s Fnet = 10 500 N upwards = (10 500)(0,01) = 105 N·s upwards 1.2 2.2 Take upwards as + Take upwards + vi = 25 m·s-1 vf = 0 m·s-1 g = -9,8 m·s-2 y = ? 25 A ball is projected vertically upwards and returns to the hand of the thrower. Take downwards as positive. 4.2 A stone is dropped from the top of a building and hits the ground after 2 s. Take downwards as positive. 4.3 A small rubber ball is dropped from a certain height, bounces twice on the ground and is caught again at its maximum height. Assume that its collision with the ground is inelastic. Take upwards as positive. â 5vi = 122,5 - 80 Take upwards as + For m motion to turning point: vf = 0 m·s-1 vi = 8,5 m·s-1 upwards g = -9,8 m·s-2 y = ? -1 â vi = 8,5 m·s upwards 3.2 vf2 = vi2 + 2g y â 0 = 8,52 + 2(-9,8) y â 19,6 y = 72,25 ANSWER Position-time 4.1 Position (m) â y = 3,69 m maximum height above the ground = 80 + 3,69 = 83,69 m 3.3 vf = vi + g t Velocity back at starting point = -8,5 m·s-1 (time symmetry) â -8,5 = 8,5 + (-9,8) t â 9,8 t = 17 Position (m) Balloon moves upwards at a constant velocity: â its displacement after t seconds: y1 = v t = 8,5 t 0 2 Time (s) Time (s) o+ 0 2 Time (s) Position-time 1 gt 2 2 1 = 8,5 t + (-9,8)t 2 2 Position (m) y2 = vi t + = 8,5 t - 4,9 t2 â Difference in distance = 0 Velocity-time ground Cell phone falls at gravitational acceleration: â its displacement after t seconds: Time (s) o+ Position-time 4.2 â t = 1,75 s y1 - y2 â 20 = 8,5 t - (8,5 t - 4,9 t2) 0 2 above building Time (s) 4.3 = 4,9 t2 Position-time 2 â t = 4,08 Position (m) 3.4 0 Velocity-time â t = 2,02 s 26 Velocity-time Time (s) 0 ground m+ 0 Time (s) Copyright © The Answer Series: Photocopying of this material is illegal MECHANICS = 42,5 Velocity (m·s-1) = 5vi - 122,5 4.1 NOTES Take upwards as + vi (cellphone) = velocity of balloon y = -80 m t = 5s g = -9,8 m·s-2 Velocity (m·s-1) 1 gt 2 2 1 â -80 = vi (5) + (-9,8)(5)2 2 y = vi t + Draw both a position-time and velocity-time graph for each of the following situations: Velocity (m·s-1) 3.1 1 Question 4 ANSWER Question 5 Question 6 25 Consider the following velocity-time graph: The following graph shows the velocity of a ball thrown vertically downwards from a height of x metres above the ground. 20 15 10 5 0 -5 1 2 4 3 5 25 Time (s) v (m·s-1) v (m·s-1) NOTES 1 -10 -15 5.1 5.2 What is the initial velocity of the object? Assume upwards is positive. How long does the object take to reach its maximum height? 5.3 At what time is the speed of the object 10 m·s-1? 5.4 Determine the position of the object relative to the ground at t = 2 s and at t = 4 s. Draw a position-time graph for the motion of the ball during the first 4 s. 5.5 C 6.1 What is the initial velocity of the ball? 6.2 What is the acceleration of the ball as it falls to the ground? 6.3 Determine the value of x. 6.4 Describe the motion of the ball for sections AB and CD of the graph. 6.5 Draw the corresponding position-time graph of the motion of the ball. t = 2,5 s to turning point 5.3 at t = 1,5 s and again at t = 3,5 s 5.4 At t = 2 s, v = 5 m·s-1 : displacement = area under graph 1 = bh + (ℓb) 2 1 = (2)(20) + (5)(2) 2 = 20 + 10 = 30 m upwards At t = 4 s, v = - 15 m·s-1 : displacement = area under graph ANSWER 25 6.1 5 m·s-1 downwards 6.2 acceleration = gradient y 2 - y1 y = = x2 - x1 x 20 15 5 0 -5 25 - 5 2-0 20 = 2 Time (s) 1 2 3 4 5 = 10 m·s-2 downwards -10 6.3 -15 vi = 5 m·s-1 area = rectangle + triangle t = 2s vf = 25 m·s-1 = 31,25 - 11,25 = 20 m upwards area = (ℓb) + Position (m) 5.5 4 Time (s) Copyright © The Answer Series: Photocopying of this material is illegal  21 bh = (5)(2) + 21 (2)(25 - 5) 20 0 + = 10 1 1 = b1h1 + b 2 h 2 2 2 1 1 = (2,5)(25) + (1,5)(-15) 2 2 MECHANICS Time (s) 4 2 v i = 25 m·s-1 upwards v (m·s-1) 5.2 D 5 0 - 20 ANSWER 5.1 B A = 10 + 20 = 30 m 27 v f 2 = v i 2 + 2a y OR â 252 = 52 + 2(10) y â 20 y = 600 y = 30 m 6.4 AB: The ball is projected vertically downwards at a velocity of 5 m·s-1. It accelerates downwards at 10 m·s-2 and hits the ground after 2 seconds with a velocity of 25 m·s-1. NOTES CD: The ball bounces back from the ground at a velocity of 20 m·s-1 and moves upwards at the same acceleration (10 m·s-2 downwards), to its turning point in the air. It reached a velocity of 0 m·s-1 at point D. (ground) 0 Position (m) 2 4 Time (s) + - 30 NB: The ball initially moved downwards (started at a negative position above the ground and moved closer to the ground (zero position)). Thereafter the ball moved upwards (in the negative direction) to a lower height from which it was thrown. MECHANICS 6.5 NOTES 1 ANSWER Copyright © The Answer Series: Photocopying of this material is illegal