1. Projectile Motion Senior Mathematics C

2. View YouTube car crash video clip here please. http://www.youtube.com/watch?v=ckAwT9AkRgE&NR=1

3. Projectile Motion3 The Problem to Solve A car travelling on a slippery mountain road (with a signed Speed Limit of 60 km/hr) goes out of control on a horizontal section of the road. Traffic accident investigators note that, on its way to the bottom of the ravine, the car clipped a rock on the mountain slope. This first mark of impact is at a horizontal distance of 104 metres from the edge of the cliff where tyre tracks indicate the car left the road (and became airborne) and at a vertical distance 122.5 metres below this point. The driver’s family want to sue the State for not maintaining the road in a safe condition and thus causing the driver’s untimely death. Do they have a case?

4. Projectile Motion4 Do you agree with this Solution? No, they do not have a case. Actually, the calculations which follow do not factor in air resistance (which would have slowed the car down a tiny bit as it travelled in its parabolic curve through the air), and the driver was probably (almost certainly, unless there was a tail wind) travelling in excess of 75 km/hr. The road signs had indicated a legal speed limit of 60 km/hr. Investigators would need to confirm that there was no appreciable wind that day.

5. Projectile Motion5 Horizontal Motion from edge of cliff to where car hit rock: s = 104 m. a = 0 (neglecting air resistance) u = u t = t s = u t + ½ a t 2 s = u t

6. Projectile Motion6 Vertical Motion from edge of cliff to where car hit rock: s = 122.5 m. (or –122.5 m, if take upwards as positive) a = 9.8 m/s 2 (or -9.8 m/s 2 if take upwards as positive) u = 0 (initially travelling horizontally, therefore no vertical component of velocity) t = t (same as for horizontal component of motion) s = u t + ½ a t 2 s = 4.9 t 2

7. Vector Representation of displacement, taking downwards as positive and where the car left the road as the origin: s = ut i + 4.9 t 2 j = 104 i + 122.5 j i j

8. Projectile Motion8 s = ut i + 4.9 t 2 j = 104 i + 122.5 j Equating corresponding components, 4.9 t 2 = 122.5-> t = 5 sec (neglect solution -5 sec as impractical here) ut = 104-> u x 5 = 104 -> u = 20.8 m/sec

9. Projectile Motion9 s = ut i + 4.9 t 2 j = 104 i + 122.5 j t = 5 sec u = 20.8 m/sec 20.8 m/sec = (20.8 / 1000) km / (1 / 3600) hr = 74.88 km/hr or approx 75 km/hr

10. Projectile Motion10 s = ut i + 4.9 t 2 j = 104 i + 122.5 j t = 5 sec u = 20.8 m/sec = (20.8 / 1000) km / (1 / 3600) hr = 74.88 km/hr or approx 75 km/hr The car was travelling 15 km/hr above the legal speed limit, and the driver would therefore be blamed for causing the accident by driving at an excessive and unsafe speed.

12. Projectile Motion12 The position vector of a projectile at time t seconds is given by: r(t) = 20t i + ( 15t – 4.9t 2 ) j metres Which of the following statements is FALSE: A. Its initial speed of projection was 25 m/sec. B. Its angle of projection was 36 o 52’ above the horizontal. C. The greatest height reached was 3.06 metres (to the nearest centimetre). D. Its range (on the horizontal plane) was 61.22 metres (to the nearest cm). E. It would have travelled exactly the same horizontal distance if it had been projected at the same speed at an angle of 53 o 08’ above the horizontal.

13. Projectile Motion13 The position vector of a projectile at time t seconds is given by: r(t) = 20t i + ( 15t – 4.9t 2 ) j metres Which of the following statements is FALSE: A. - B. - C. The greatest height reached was 3.06 metres (to the nearest centimetre). D. - E. - The greatest height reached was actually 11.5 metres.

14. Projectile Motion14 The position vector of a projectile at time t seconds is given by: r(t) = 20t i + ( 15t – 4.9t 2 ) j metres v(t) = 20 i + ( 15 – 9.8 t) j At greatest height when vertical component of velocity is zero. i.e. when 15 – 9.8 t = 0 t = 15 / 9.8 Therefore Greatest Height reached = 15 x ( 15 / 9.8 ) – 4.9 x ( 15 / 9.8 ) 2 = 11.5 metres (and not 3.06 m)

15. Projectile Motion15 The position vector of a projectile at time t is r(t) = 10t i + ( 19.6t – 4.9t 2 ) j (metres) where i is horizontal and j is vertically upwards. What is the maximum height reached by this projectile. A. 0 metres B. 4.9 metres C. 9.8 metres D. 19.6 metres E. 39.6 metres

16. Projectile Motion16 The position vector of a projectile at time t is r(t) = 10t i + ( 19.6t – 4.9t 2 ) j (metres) where i is horizontal and j is vertically upwards. What is the maximum height reached by this projectile. A. - B. - C. - D. 19.6 metres E. -

17. Projectile Motion17 The position vector of a projectile at time t is r(t) = 10t i + ( 19.6t – 4.9t 2 ) j (metres) where i is horizontal and j is vertically upwards. What is the maximum height reached by this projectile. v(t) = 10 i + ( 19.6 – 9.8 t ) j Max Ht when 19.6 – 9.8 t = 0 i.e. t = 19.6 / 9.8 = 2 seconds Max Ht = 19.6 x 2 – 4.9 x 2 2 = 19.6 metres

18. Projectile Motion18 For projection on a horizontal plane, the maximum range is attained when the angle of projection is 45 degrees above the horizontal. To achieve the maximum horizontal displacement from the bottom of a 100 metre high cliff, an object thrown from the top of the cliff at a speed of 20 m/sec should be thrown at an angle to the horizontal of: A. 0 o B. 22.5 o C. 45 o D. 67.5 o E. 90 o

19. Projectile Motion19 For projection on a horizontal plane, the maximum range is attained when the angle of projection is 45 degrees above the horizontal. To achieve the maximum horizontal displacement from the bottom of a 100 metre high cliff, an object thrown from the top of the cliff at a speed of 20 m/sec should be thrown at an angle to the horizontal of: A. - B. 22.5 o C. - D. - E. -

20. Projectile Motion20 For projection on a horizontal plane, the maximum range is attained when the angle of projection is 45 degrees above the horizontal. To achieve the maximum horizontal displacement from the bottom of a 100 metre high cliff, an object thrown from the top of the cliff at a speed of 20 m/sec should be thrown at an angle to the horizontal of between 0 and 45 degrees. (The maximum range will occur when as if the object was thrown at ground level at an angle of 45 o, and reached a height of 100 metres on the way to its greatest height.) 45 o 22.5 o 100 m high cliff

21. Projectile Motion21 A bullet is fired from a gun at a speed of 160 m/sec horizontally at a target 30 metres away. The gun barrel was aimed directly at the bullseye. How far below the bullseye will the bullet hit the target? A. 3 cm B. 17 cm C. 30 cm D. 1.7 m E. 3 m

22. Projectile Motion22 A bullet is fired from a gun at a speed of 160 m/sec horizontally at a target 30 metres away. The gun barrel was aimed directly at the bullseye. How far below the bullseye will the bullet hit the target? A. - B. 17 cm C. - D. - E. -

23. Projectile Motion23 A bullet is fired from a gun at a speed of 160 m/sec horizontally at a target 30 metres away. The gun barrel was aimed directly at the bullseye. How far below the bullseye will the bullet hit the target? Time Taken = 30 / 160 sec. Vertical Displacement = ½ x 9.8 x ( 30 / 160 ) 2 = 0.172265625 metres (or approximately 17 cm)

24. Projectile Motion24 A bullet is fired from a gun at a speed of 160 m/sec horizontally at a target 30 metres away. The gun barrel was aimed directly at the bullseye. For the bullet to hit the target, at what angle to the horizontal should the gun be aimed? A. 0.1 o B. 0.2 o C. 0.33 o D. 0.67 o E. 1 o

25. Projectile Motion25 A bullet is fired from a gun at a speed of 160 m/sec horizontally at a target 30 metres away. The gun barrel was aimed directly at the bullseye. For the bullet to hit the target, at what angle to the horizontal should the gun be aimed? A. - B. - C. 0.33 o D. - E. -

26. Projectile Motion26 A bullet is fired from a gun at a speed of 160 m/sec horizontally at a target 30 metres away. The gun barrel was aimed directly at the bullseye. For the bullet to hit the target, at what angle to the horizontal should the gun be aimed? Horizontally, 160 cos  t = 30 Vertically, 160 sin  t – 4.9 t 2 = 0 i.e. t = ( 160 sin  ) / 4.9 and ( 160 2 cos  sin  ) / 4.9 = 30 Using sin 2  = 2 sin  cos  sin 2  = 0.011484375 giving 2  = 0.658 o and  = 0.33 o (approx.)

28. Projectile Motion28 Student Investigation In Numb3rs TV episode “Convergence”, FBI Agents Colby and David are trying to locate a bullet that was fired as a warning shot over two carjack victim’s heads. They turn to Charlie for advice about where to look for the bullet. Charlie uses parametric equations and a Graphics Calculator to simulate the trajectory of the bullet. After taking all of the factors into account, Colby and David are able to locate the missing bullet. David and Colby are trying to find the location of a bullet that was fired from a 9mm pistol. From witness reports, they estimate that the gun was five feet (about 1.5 metres) off the ground when it was shot. The shooter was standing in a large, level field and fired one shot over the victim’s heads to scare them. David and Colby estimate that the angle of elevation of the gun when it was fired was 10 degrees. They know that a 9mm pistol can launch a bullet with an initial velocity of 455 feet per second (approx. 140 metres/second) (after air resistance has been taken into account). If the victims were standing 45 feet (approximately 14 metres) away from the shooter when he shot the gun, about how far did the bullet go over the victims’ heads? (Assume that the victims were on average, 6 feet (approximately 1.85 metres) tall.) Approximately what was the maximum altitude reached by the bullet? About how long was the bullet in the air? About how many metres did the bullet travel in the horizontal direction?

30. Projectile Motion30 Homework Exercise A truck driver, descending a mountain road at an incline of 1 in 4 (14 degrees below the horizontal), where the legal speed limit is 50 kilometres / hour, failed to make a corner and drove his vehicle off the road, over the edge of a cliff. Accident scene investigators noted that the vehicle clipped the top of a pine tree during its mid-air flight. The broken branch was measured as 100 metres horizontally distant and 155 metres vertically below the point the vehicle left the edge of the cliff. You are called in as an expert witness to testify as to the speed at which the truck was travelling prior to the accident. (A passenger in the vehicle was killed in the incident, and the driver, who claims he was travelling at 48 km/hr at the time, has been charged with dangerous driving causing death.) Are you a witness for the prosecution or the defence? What assumptions have you made in arriving at your estimate of the vehicle’s speed and how confident are you of the accuracy of this estimate?

32. References / Sources http://www.youtube.com/watch?v=ckAwT9AkRgE&NR=1 http://education.ti.com/educationportal/activityexchange/Activity.do ?cid=US&aId=6181 Senior Mathematics C