Physics Unit 2: Projectile Motion

Transcription

1 Physics Unit 2: Projectile Motion Jan 31 10:07 AM What is a projectile? A projectile is an object that is launched, or projected, by some means and continues on its own inertia. The path of the projectile is called the trajectory. Projectile motion can be described by two components; horizontal & vertical. Jan 31 10:08 AM 1

2 The horizontal component (v x ) can be described with a constant velocity. A dot diagram would look like this We can describe this motion as v x =d x or x t t The object covers an equal distance in equal intervals of time. Jan 31 10:12 AM The vertical component (v y ) behaves like a freely falling body. It is affected by gravity and accelerates downward. A dot diagram for v y would look like this. Jan 31 10:20 AM 2

3 Adding the vectors v x and v y together shows us that the path of the projectile's motion is a parabola Jan 31 10:25 AM Example: A cannon ball continually falls beneath a line drawn to show its path in the absence of gravity until it finally strikes the ground. The vertical distance it falls beneath any point on the dotted line is the same vertical distance it would fall if it were dropped from rest and had been falling for the same amount of time. d = 1/2gt 2 45m path without gravity 5m 20m 1s 2s 3s Shoot a projectile in the air and if there is no gravity it will be on that path at a distance v/d. But instead it is directly under that point. How far? 4.9t 2 (1/2gt 2 = 4.9t 2 ) Jan 31 10:47 AM 3

4 If the projectiles launched from different altitudes than the target, there will be different horizontal ranges. Vy = 0 m/s at the top of the trajectory t up = t down Horizontal distance x = v x t or x = v i cosθt = v i (cosθ)t Vertical distance y = v y t +1/2gt 2 v y = v i sinθ y = ( v i sinθ)t + 1/2gt 2 Final velocity v fy = v iy + gt v f 2 = v iy + gd v fy 2 = v iy 2 +2g y g = -9.8m/s 2 Jan 31 11:16 AM Horizontal Launch Jan 31 1:40 PM 4

5 Using Kinematic Equations with Projectiles Constant velocity in the x-direction x = v ix t x = (v ι cosθ)t Velocity in the y-direction is affected by gravity, a y = -9.8m/s 2 v fy = v iy +a y t y = v iy t + 1/2a y t 2 v fy 2 = v iy 2 +2a y y At any point along a projectiles path, if we know v x + v y, then we can find the resultant velocity and angle by: v 2 = v x 2 + v y 2 and θ = tan -1 (v y /v x ) Jan 31 1:13 PM Horizontal Launch For a horizontal launch: We know a x = 0 m/s 2 a y = -9.8m /s 2 o θ = 0 v ix = v i cosθ = v i cos0 = v i v iy = v i sinθ = v i sin 0 = o o You can solve for most things dealing with a horizontal launch using these two equations So x = v ix t becomes x =v i t y = v iy t + 1/2a y t 2 becomes y = 1/2a y t 2 Jan 31 12:42 PM 5

6 Horizontal Launch Sometimes a problem may ask for the velocity when an object hits the ground. In this case you must find v f. Remember that v f 2 = v fx 2 + v fy 2 or v f = v fx + v fy Since velocity is constant in the x direction, v fx = v ix = v i In the y direction velocity is changing! 0 v fy = =v iy +a y t so v fy = -9.8t Jan 31 1:01 PM A rock is thrown horizontally from a 100m high cliff. It strikes the ground 90.0m from the base of the cliff. At what speed was it thrown? (Hint: find the time of falling first!) a x = 0m/s 2 a y = 9.8m/s 2 v iy = 0m/s d y = 100 m d x = 90m solve for t 100m? 9.8 m/s 2 90 m d x = v ix t d y = 1/2a y t 2 90m = v ix (4.52s) 2d y /a y = t 2 v ix = 90m/4.52s 2(100m/s)/9.8m/s 2 = t 2 v ix = 19.9 m/s t = 4.52 s substitute and solve for v ix Jan 31 12:04 PM 6

7 Practice; Problems for projectiles launched horizontally Solve Workbook Problems D pp. 23 & 24 Answers: 1. the river is 5.98 m wide m/s. No, 172m/s is much larger than 100m/s m m m m/s m/s m/s; 64 degrees below horizontal Jan 31 12:21 PM Physics: Motion in 2 Dimensions Projectiles Launched at an Angle Jan 31 12:38 PM 7

8 For an object launched at an angle: v y = 0 m/s at max height t up θ x y tdown For most problems you are given v i and θ. You are asked to find the (1) total time the object is in the air, (2) the range, and (3)the maximum height the object reaches. Jan 31 12:41 PM 1). total time the object is in the air: v fy = v iy +a y t up where v iy = v i sinθ 0 = v i sinθ - 9.8t up when it hits the ground t up = v i sinθ 9.8 Total time in the air = 2 t up Jan 31 1:50 PM 8

9 2) Range: x = v ix t where v ix = v i cosθ O = (v i sin θ)t Use total time! If no time is given use: x = v i 2 sin2θ g Jan 31 2:21 PM 3). maximum height: y = v iy t = 1/2a y t 2 = (v i sinθ)(t) + 1/2(-9.8m/s 2 )(t) 2 Use time at max height! (t up ) Jan 31 2:31 PM 9

10 θ v i v ix v iy y motion: in the y direction, objects slow as they go up ( ), stops at the top, increases speed as it falls( ). Because the speed is changing in the y-direction, we know that it is accelerating! It is getting a push from gravity in the y direction (a = -9.8m/s 2 ) x-motion: once an object is launched there is no force (push) acting on it in the x-direction. Acceleration in the x-direction is zero. Velocity in the x direction is constant. Jan 31 2:39 PM Example: A golf ball is hit and leaves the tee with a velocity o of 25.0 m/s at 35.0 with respect to the horizontal. What is the horizontal displacement of the ball? Find t first! o 25.0 m/s 35.0 v = 25m/s θ = 35.0 v x = v cosθ x =? x = v x t = (25 m/s)(cos 35.0) = 20.5 m/s x v y = (25 m/s)(sin 35.0) = 14.3 m/s v y = v y T + 1/2gT 2 0 = (14.3 m/s)t + 1/2(-9.8 m/s 2 )T 2 T T = v x t =(20.5m/s)(2.9s) = m 0 = 14.3m/s - 4.9m/s 2 T T = 2.9s Jan 31 3:00 PM 10

11 Practice; Problems for projectiles launched at an angle Solve Workbook Problems E pp ). v i = 45.8 m/s 2). v = 68.2 m/s 3). v i = 16.6 m/s 4 a) v i = 20.8 m/s b). the brick's max height is 11.0 m c) m 5). y max = 4.07 m 6). θ = 60 o 7). x = 6.73 m 8). the flea's max height is 3.2 cm Jan 31 3:43 PM 11