Copyright Sautter 2003 Motion in Two Dimension Projectiles

Copyright Sautter 2003

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(Sy and Sx), velocities (Vyand Vx) and accelerations (Ay and Ax) ( g, of course, refers to gravity)

CLICK HERE JUST AFTER FIRING Sy =0, Sx = 0 Vy = +max Vx = constant Ay = g, Ax =0 AT HIGH POINT Sy = max, Sx = ½ max Vy = 0, Vx = constant Ay = g, Ax =0 JUST BEFORE LANDING Sy = 0, Sx = max Vy = -max Vx = constant Ay = g, Ax =0

A Y COMPONENT X COMPONENT Y COMPONENT C X COMPONENT B Y COMPONENT

Vy = 0 Sy Vy inst = slope of tangents Time Ay inst = slope of tangents Vy Ay Time Ay inst = - 9. 8 meters / sec 2

Path of object Without gravity Object Projected at Angle Vo Actual projectile path Distance fallen due to gravity (1/2 g t 2) Vertical height if gravity did not act on the object (Vo sin t) Sy = actual height of object Sx Sy = height without effects of gravity – distance fallen due to gravity Sy = Vo sin t + ½ g t 2 (the value of gravity is negative) Horizontal distance traveled by projectile Sx = Vo cos t

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! Vy = Vo sin + gt Vx = Vo cos

Projectile Vx > 0 Constant velocity Dropped Object Vx = 0 Accelerated by gravity 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 !

Vy = 0 Vy = + max Vy = - max Horizontal component of projectile motion (Constant velocity – no acceleration) Vertical component of projectile motion (Accelerated by gravity)

Vertical Displacement Horizontal Displacement Vertical Velocity Horizontal Velocity Vertical Acceleration Horizontal Acceleration

Maximum Height hmax Maximum Horizontal Distance (Range) range

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 !

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? Vo = 12 m/s • “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 30 m 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. 47 sec (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 Vx = 12 (cos 0) 2. 47 = 29. 6 meters

Solving Projectile Problems A ball is thrown upward at a 300 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? 300 Vo = 12 m/s • 300 m • The object is projected at + 300. Vo = 12 ft/s and we are asked in part (a) to find the time when Sy = - 300 ft. In part (b) we are asked to find the horizontal distance (Sx). (a) using the equation shown we insert – 300 for Sy, 12 for Vo 300 for the angle and – 32 ft/s 2 for gravity (ENG) and get – 300 = 12(sin 300 ) 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 Vx = 12 (cos 300) 5. 46 = 56. 7 feet

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? hmax R • Vo = 20, the angle = 450 and in part (a) we are asked to find hmax • In part (b) we are asked to find the range ( R) (a) using the equation shown we insert 20 for Vo 450 for the angle and – 32 ft/s 2 for gravity (ENG) and get hmax = - ((20)2 (sin 450)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 450)) / - 32 = 12. 5 feet

Solving Projectile Problems A ball is throw downward at an angle of 300 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? -300 Vo = -10 m/s 40 m Downward = - 300 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 Vx value and the Vy value and add them using vectors. Using the given equations, Vo = 10 m/s, g = - -9. 8 m/s 2 (MKS) and Sy = - 40 m we must first find time (t). -40 = 10 sin (-300) 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, Vy = 10 (sin – 300) + (-9. 8) 1. 78, Vy = -22. 5 m/s To find the horizontal velocity, Vx = 10 (cos – 300) 1. 78 = +15. 4 m/s

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). Vx = +15. 4 m/s Vy = -22. 5 m/s Vresulting • To find V resulting we use vector addition (the Pythagorean Theorem), Vr = ((-22. 5)2 + (15. 4)2)1/2 , Vr = 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.

Solving Projectile Problems • 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? Vo = 20 V = - 5, t = ? 450 • Vy = -5 ft/s, Vo = 20 ft/s and the angle of projection = 450. Using the given equation, -5 = 20 (sin 450) + (- 32 ) t • t = 0. 598 seconds Check - At the highest point, Vy =0 and 0 = 20 (sin 450) + (-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!

A ball is thrown horizontally at 12 m/s. (a) 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 Hint: use The vector Sum of Vx + Vy A ball is thrown at 400 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 Gravity is its acceleration After 0. 50 seconds ? It is -9. 8 m /s 2 (A) 2. 6 m/s (B) 3. 1 m/s (C) 4. 9 m/s (D) 7. 5 m/s (MKS units) Hint: maximum What must be the original velocity of a projectile horizontal in order to reach a target 90 km away ? distance occurs (A) 320 m/s (B) 939 m/s (C) 882 m/s (D) none at an angle of 450 A ball is thrown at a 300 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 Click Here For answers

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.