Projectile Motion 2 Launch Angles (symmetrical and asymmetrical trajectories) Physics 12 Comprehension Check An arrow is launched from a cliff with an initial velocity of 25m/s. If the cliff is 25m high determine: The time in the air How far from the base of the cliff it hits How fast it is going when it hits the ground Projectile Diagram – Launched at an Angle Projectiles Launched at an Angle Both vi and vf have x and y components Still follow a parabolic trajectory Symmetrical Trajectories When you launch a projectile from the same vertical height that it lands dy = 0 Think back to physics 11… what do you know about the speed/velocity of an object that goes the same distance up as it goes down? viy = -vfy Symmetrical Trajectories What about the overall time and the time to its peak? Time to peak/vertex = ½ time in air What do you think about the angle it is launched at and the angle it lands at? θi = θf What about the x-direction? Is there any force acting on the projectile once it is launched? Acceleration? No force acting on the projectile once launched except gravity, which only affects the y-direction. What can we say about vix and vfx? If ax = 0, then vix and vfx Hint… Remember: To find the initial velocity in the x and y direction, you must use trigonometry! X direction: cosine Y direction: sine Remember that at its peak, an object’s y velocity is 0 because it is changing from a positive velocity to a negative velocity (grade 11). Example 1 A player kicks a football with an initial velocity of 29 m/s at 69.0°. Assuming it lands at the same height from which it was kicked, determine the time of flight. Projectiles Launched at an Angle A golfer uses a club that launches a golf ball at a 15° angle at a speed of 45m/s. Determine the following: The time the golf ball is in the air The horizontal distance the ball travels The velocity as it strikes the ground The maximum height the ball attains Start with the position in the y equation Include the initial velocity in the y term Solve for when the position in the y is equal to zero 2 gt dy viyt 2 0 4.91t 2 45 sin( 15)t 4.91t 2 45 sin( 15)t 4.91t 45 sin( 15) t 2.4 s Use the time from the previous question and the position in the x equation Solve for the range (horizontal distance) d x vixt d x 45 cos(15)( 2.4 s ) 2 d x 1.0 x10 m Solve for the velocity in the x and y direction v fx v fx v fx v fy v fy v fy vix 45 cos(15) 44 gt viy 9.81(2.4 s ) 45s sin( 15) 12m / s Use Pythagorean Theorem and Trig to solve for final velocity v x 44m / s v y (2.4 s ) 12m / s v (44m / s ) (12m / s ) 2 v 45m / s 12m / s tan 44m / s o 15 v 45m / s,15o 1 2 Max height will occur when y velocity is equal to zero. Solve for time and then sub into y position equation v fy gt viy 0 9.81t 45 sin( 15) t 1.2s 2 gt d fy viyt 2 d fy 4.91m / s 2 (1.2s ) 2 45 sin( 15)(1.2s ) d fy 6.9m Comprehension Check A cannonball is launched with an initial velocity of 25m/s at an angle of 32°. How long is it in the air? What is the horizontal distance that the cannonball travels? What is the maximum height? With what velocity does it strike the ground? At 56m, there is a 2.0m high wall; does the cannonball clear the wall? Comprehension Check 2.7s 57m 8.9m 25m/s, 32° No Practice Problems Page 536 Questions 1-8 Page 549 Questions 13-14 Asymmetrical Trajectories When the launch height is different from the landing height Launching a projectile at an angle from the top of a hill Launching a projectile from ground level into water Launching a projectile from the ground and it hitting a hill Final and initial y velocities will not be the same Final and initial x velocities will be the same Time to the vertex/peak will not be ½ time Vertical velocity will still be 0 at peak Things to remember… You will need to use the quadratic equation and… If the projectile lands HIGHER than the launch site: There will be 2 positive roots but the first root will be the time it takes to get to that height on the way up (so use the other value) If the projectile lands LOWER than the launch site: There will be a 1 positive root (the negative root will be before the launch happened) Example 1 A golfer hits the golf ball off the tee, giving it an initial velocity of 32.6 m/s at an angle of 65’ with the horizontal. The green where the golf ball lands is 6.30m higher than the tee. Neglect air friction. Find: The time the golf ball is in the air (5.80s) How far it travels horizontally (79.9m) The velocity of the ball just before it hits the ground (30.6 m/s [E63.3S]) Try this… A golfer strikes a ball with a velocity of 470 m/s at 35.0’. If the ball lands on the fairway 3.50m below the tee, find the Time of flight (5.63s) Range (217 m) Max height (37.0 m) Velocity at end of flight (47.7 m/s [E 36.2’S]) Try this… A golfer stands on an elevated tee that is 6.50m higher than the green. If the ball next to the cup (183m away) and the ball is hit at a 40.0° angle, determine: The speed at which the ball left the club (41.8 m/s) The time the golf ball is in the air (5.71s) The maximum height the ball attains (36.8m) Practice Page 543, Questions 9 to 12

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