The document discusses projectile motion in two dimensions, detailing how objects move both vertically and horizontally under the influence of gravity. It explains key concepts, equations, and examples related to projectile motion, such as the independence of horizontal and vertical components and how to calculate height and range of projectiles. Additionally, it contrasts ideal scenarios with real-life effects like air resistance and terminal velocity.

1. PROJECTILES Motion in Two Dimensions Copyright Sautter 2011

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4. Motion in Two Dimension - Projectiles Projectile motion involves object that move up or down and right or left simultaneously. A ball thrown into the air at an angle is a common example as is water sprayed from a garden hose or a bullet fired at a target. Any object once released from its accelerating force is accelerated downward by gravity at all times during its flight. This causes a continual variation in its vertical velocity while its horizontal velocity remains unaffected. It discussing projectile motion, the horizontal and vertical components are treated independently. The combined effect of these two motions on the object give its parabolic flight path. When objects are projected from a specific horizontal elevation and land at the same horizontal elevation, the parabolic path of the object is symmetrical. The next slide shows a projectile path and key points during the flight in terms of displacements(S y and S x ), velocities (V y and V x ) and accelerations (A y and A x ) ( g, of course, refers to gravity)

5. Projectile Motion JUST AFTER FIRING S y =0, S x = 0 V y = +max V x = constant A y = g, A x =0 AT HIGH POINT S y = max, S x = ½ max V y = 0, V x = constant A y = g, A x =0 JUST BEFORE LANDING S y = 0, S x = max V y = -max V x = constant A y = g, A x =0 CLICK HERE

6. A X COMPONENT Y COMPONENT X COMPONENT Y COMPONENT B Y COMPONENT X COMPONENT C Vector Components

7. S y V y inst = slope of tangents V y Time V y = 0 A y inst = slope of tangents A y Time A y inst = - 9.8 meters / sec 2 Projectile Motion (vertical component) Time

8. V o S x Path of object Without gravity S y = actual height of object Distance fallen due to gravity (1/2 g t 2 ) Vertical height if gravity did not act on the object (V o sin  t)  S y = height without effects of gravity – distance fallen due to gravity S y = V o sin  t + ½ g t 2 (the value of gravity is negative) Horizontal distance traveled by projectile S x = V o cos  t Vertical & Horizontal Displacements of a Projectile Actual projectile path Object Projected at Angle 

9. Water spraying from a hose is a common example of projectile motion The vertical motion of the water is accelerated by gravity. The horizontal motion is constant velocity and is unaffected by gravity! Velocity Components of Projectile Motion V y = V o sin + gt V x = V o cos  

10. Vertical displacements are equal for projectiles and dropped objects however horizontal displacements are greater for projectiles. Vertical acceleration for both dropped objects and projectiles is that of gravity (-32 ft/s 2 , -9.8 m/s 2 ). They both hit the ground at the exact same time but, of course, the projectile is further away ! Accelerated by gravity Projectile V x > 0 Constant velocity Vertical & Horizontal Motion of a Projectile vs. a Dropped Object Dropped Object V x = 0

11. Vertical & Horizontal Components of Motion during Projectile Flight V y = + max V y = - max V y = 0 Horizontal component of projectile motion (Constant velocity – no acceleration) Vertical component of projectile motion (Accelerated by gravity)

12. Projectile Motion Equations Vertical Displacement Horizontal Displacement Vertical Velocity Horizontal Velocity Vertical Acceleration Horizontal Acceleration a = g y a = o x y o S = V sin t + 1/2 g t 2  V = V cos x o  V = V sin + g t y o  S = V cos t x o 

13. Special Situation Projectile Equations Maximum Height Maximum Horizontal Distance (Range) h = - V sin max o 2 2 2 g -----------------  R = - V sin 2  2 ----------------- o g h max range

14. When an object is dropped at the exact same time a projectile is fired at the falling object, aiming directly at the object always insures a direct hit. Why ? Because the object and the projectile once fired, are in both in free fall ! Projectiles in Free Fall

15. Solving Projectile Problems A ball is thrown horizontally at 12 m/s from a building 30 meters high. (a) How long will is be in the air? (b) How far from the base of the building will it land? “ horizontally” means the angle of projection is 0 degrees In part (a) we are asked to find time when the vertical distance is – 30 meters (negative means below the point of release). In part (b) we are asked to find horizontal distance (a) Using the equation shown and inserting –30 for vertical distance, 12 for the original velocity, 0 degrees for the angle and – 9.8 for gravity (MKS) we get: -30 = 12 (sin 0)t + ½ (-9.8)t 2 , solving for t gives 2.47sec ( b) Using the equation shown and inserting 12 for the original velocity, 0 degrees for the angle and 2.4 seconds for time found in part (a) we get V x = 12 (cos 0) 2.47 = 29.6 meters Equations to be Used V o = 12 m/s 30 m y o S = V sin t + 1/2 g t 2  S = V cos t x o 

16. The object is projected at + 30 0 . V o = 12 ft/s and we are asked in part (a) to find the time when S y = - 300 ft. In part (b) we are asked to find the horizontal distance (S x ). Solving Projectile Problems A ball is thrown upward at a 30 0 angle, at 12 ft/s from a building 300 feet high. (a)How long will is be in the air? (b) How far from the base of the building will it land? (a) using the equation shown we insert –300 for S y , 12 for V o 30 0 for the angle and – 32 ft/s 2 for gravity (ENG) and get –300 = 12(sin 30 0 ) t + ½ (-32) t 2 which gives –300 = 6 t – 16 t 2 or 16 t 2 –6 t - 300 = 0 (a quadratic). Using a = +16, b = - 6 and c = - 300 we insert these values in the quadratic equation find t = -3.21 or + 5.46 The negative t means the ball hits the ground before it is thrown in is therefore obviously wrong. t = +5.46 sec (b) Using the equation shown and the time value from part (a) we get V x = 12 (cos 30 0 ) 5.46 = 56.7 feet Equations to be Used V o = 12 m/s 300 m 30 0 y o S = V sin t + 1/2 g t 2  X + X + = y 2 a b c x = - b b - 2 a + - 2 ------------------------- / \ ----------------------------- 4 a c S = V cos t x o 

17. Solving Projectile Problems A projectile is shot on level ground at a 45 degree angle with a velocity of 20 ft/s. (a) How high will it travel? (b) How far will it go? V o = 20, the angle = 45 0 and in part (a) we are asked to find h max In part (b) we are asked to find the range ( R) Equations to be Used (a) using the equation shown we insert 20 for V o 45 0 for the angle and – 32 ft/s 2 for gravity (ENG) and get h max = - ((20) 2 (sin 45 0 ) 2 )/ (2 (-32)) which gives (400 x (0.707 )2 ) / -64 . Maximum height = 3.13 feet. (b) Using the equation shown we get R = - ((20) 2 x (sin 2 x 45 0 )) / - 32 = 12.5 feet h max R h = - V sin max o 2 2 2 g -----------------  R = - V sin 2  2 ----------------- o g

18. Solving Projectile Problems A ball is throw downward at an angle of 30 0 with a velocity of 10 m/s from a building 40 meters high. What is the velocity of the ball when it hits the ground? Downward = - 30 0 We are asked to find the velocity (a vector quantity - direction counts) when the object hits the ground. At this point it is moving both horizontally and vertically. We will first find the V x value and the V y value and add them using vectors. Using the given equations, V o = 10 m/s, g = - -9.8 m/s 2 (MKS) and S y = - 40 m we must first find time (t). -40 = 10 sin (-30 0 ) t + (-9.8) t 2 , 9.8 t 2 + 5 t – 40 = 0. Solving the quadratic formula with a = 9.8, b = 5 and c = -40 we get t = - 2.29 or + 1.78 seconds. Of course, the + 1.78 is the correct answer. To find the vertical velocity, V y = 10 (sin – 30 0 ) + (-9.8) 1.78, V y = -22.5 m/s To find the horizontal velocity, V x = 10 (cos –30 0 ) 1.78 = +15.4 m/s V o = -10 m/s 40 m -30 0 V = V sin + g t y o  V = V cos x o  y o S = V sin t + 1/2 g t 2 

19. Previous Problem (continued) When the object hits the ground it is moving both vertically under the influence of gravity and horizontally at constant velocity (unaccelerated). To find V resulting we use vector addition (the Pythagorean Theorem), V r = ((-22.5) 2 + (15.4) 2 ) 1/2 , V r = 27.3 m/s Vector values such as velocity requires a direction also. Using the inverse tangent we get, tan –1 (15.4 / 22.5 ) = 34.4 0 The velocity at which the object hits the ground is 27.3 m/s and an angle of 34.4 degrees below the horizontal. V y = -22.5 m/s V x = +15.4 m/s V resulting

20. A projectile is shot on level ground at a 45 degree angle with a velocity of 20 ft/s. When will its velocity be –5 ft/s? Solving Projectile Problems V y = -5 ft/s, V o = 20 ft/s and the angle of projection = 45 0 . Using the given equation, -5 = 20 (sin 45 0 ) + (- 32 ) t t = 0.598 seconds Check - At the highest point, V y =0 and 0 = 20 (sin 45 0 ) + (-32) t = 0.442 seconds at the highest point and when descending t must be larger than 0.442 and smaller than 0.884 (2 x 0.442), the time when the object lands. The answer is between these two values! V o = 20 45 0 V = - 5, t = ? V = V sin + g t y o 

21. Now it's time for you to try some problems on your own ! The problems are similar to the ones which have been solved so look back and review the appropriate problem if you get stuck !

22. A ball is thrown horizontally at 12 m/s. What is its velocity after 2.0 seconds? (A) 12 m/s (B) 19.6 m/s (C) 23 m/s (D) 31.6 m/s Click Here For answers A ball is thrown at 40 0 above the horizontal at 4.0 m/s. What is its horizontal velocity after 0.50 seconds? (A) 2.6 m/s (B) 3.1 m/s (C) 3.4 m/s (D) 5.5 m/s What is the vertical velocity in the previous problem After 0.50 seconds ? (A) 2.6 m/s (B) 3.1 m/s (C) 4.9 m/s (D) 7.5 m/s What must be the original velocity of a projectile in order to reach a target 90 km away ? (A) 320 m/s (B) 939 m/s (C) 882 m/s (D) none Gravity is its acceleration It is -9.8 m /s 2 (MKS units) A ball is thrown at a 30 0 angle at 10 m/s. How far on the horizontal will it land ? (A) 9.8 m (B) 0.88 ft (C) 8.8 m (D) 10.2 m Hint: use The vector Sum of V x + V y Hint: maximum horizontal distance occurs at an angle of 45 0

23. Real Projectile Motion In actual free fall situations objects reach a terminal velocity and do not accelerate at a constant rate throughout the entire fall. The buoyant force to the air acts to reduce the acceleration caused by gravity. The degree to which this retarding force acts, depends on several variables. As the density of the air increases, buoyancy increases. Also, increased surface area increases the effects of buoyancy and causes a decrease in terminal velocity. The shape and aerodynamic properties of the object also effect terminal velocity. Most importantly, as the velocity of the falling body increases, the force opposing the fall increases, thus at a specific velocity the force of gravity and opposing force become equal and terminal velocity is reached.

24. Ideal vs. Real Projectile Paths Projectile Path without Air Resistance Projectile Path with Air Resistance

25. The End