projectile.notebook February 01 and 2, 2010 Physics Unit 2: Projectile Motion Jan 3110:07 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 3110:08 AM 1 projectile.notebook February 01 and 2, 2010 The horizontal component (vx) can be described with a constant velocity. A dot diagram would look like this We can describe this motion as vx =dx or ∆x t t The object covers an equal distance in equal intervals of time. Jan 3110:12 AM The vertical component (vy) behaves like a freely falling body. It is affected by gravity and accelerates downward. A dot diagram for vy would look like this. Jan 3110:20 AM 2 projectile.notebook February 01 and 2, 2010 Adding the vectors vx and vy together shows us that the path of the projectile's motion is a parabola Jan 3110:25 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 = 1/2gt2 ou ity av r tg th pa th wi 45m 20m 5m 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? 4.9t2 (1/2gt2 = 4.9t2) Jan 3110:47 AM 3 projectile.notebook February 01 and 2, 2010 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. 75 t up = t down 60 45 30 15 Horizontal distance ∆x = vxt or ∆x = vicosθt = vi (cosθ)t Final velocity vfy = viy + gt vf2 = viy + gd vfy2 = viy2 +2g∆y Vertical distance ∆y = vyt +1/2gt2 vy = visinθ ∆y = ( visinθ)t + 1/2gt2 g = -9.8m/s2 Jan 3111:16 AM Horizontal Launch Jan 311:40 PM 4 projectile.notebook February 01 and 2, 2010 Using Kinematic Equations with Projectiles Constant velocity in the x-direction ∆x = vixt ∆x = (vιcosθ)t Velocity in the y-direction is affected by gravity, ay = -9.8m/s2 vfy = viy +ayt ∆y = viyt + 1/2ayt2 vfy2 = viy2 +2ay∆y At any point along a projectiles path, if we know vx + vy, then we can find the resultant velocity and angle by: v2 = vx2 + vy2 and θ = tan-1(vy/vx) Jan 311:13 PM Horizontal Launch For a horizontal launch: We know ax = 0 m/s2 ay = -9.8m /s2 o θ=0 1 o vix = vicosθ = vicos0 = vi 0 o You can solve for most things dealing with a horizontal launch using these two equations viy = visinθ = visin 0 = 0 So ∆x = vixt becomes ∆x =vit ∆y = viyt + 1/2ayt2 becomes ∆y = 1/2ayt2 Jan 3112:42 PM 5 projectile.notebook February 01 and 2, 2010 Horizontal Launch Sometimes a problem may ask for the velocity when an object hits the ground. In this case you must find vf. Remember that vf2 = vfx2 + vfy2 or vf = vfx + vfy Since velocity is constant in the x direction, vfx = vix = vi In the y direction velocity is changing! 0 vfy = =viy +ayt so vfy = -9.8t Jan 311:01 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!) sol dx = vixt 90m = vix(4.52s) vix = 90m/4.52s vix = 19.9 m/s 9.8 m/s2 100m ? ax = 0m/s2 ay = 9.8m/s2 viy = 0m/s dy = 100 m dx = 90m ve fo rt 90 m dy = 1/2ayt2 2dy/ay = t2 2(100m/s)/9.8m/s2 = t2 t = 4.52 s substitute and solve for vix Jan 3112:04 PM 6 projectile.notebook February 01 and 2, 2010 Practice; Problems for projectiles launched horizontally Solve Workbook Problems D pp. 23 & 24 Answers: 1. the river is 5.98 m wide 2. 172 m/s. No, 172m/s is much larger than 100m/s 3. 170 m 4. 0.735 m 5. 39 m 6. 64 m/s 7. 9.8 m/s 8. 6.88m/s; 64 degrees below horizontal Jan 3112:21 PM Physics: Motion in 2 Dimensions Projectiles Launched at an Angle Jan 3112:38 PM 7 projectile.notebook February 01 and 2, 2010 For an object launched at an angle: vy = 0 m/s at max height td ∆y t up ow n θ ∆x For most problems you are given vi 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 3112:41 PM 1). total time the object is in the air: vfy = viy +aytup where viy = visinθ 0 = visinθ - 9.8tup tup = visinθ 9.8 Total time in the air = 2 tup when it hits the ground Jan 311:50 PM 8 projectile.notebook February 01 and 2, 2010 2) Range: ∆x = vixt where vix = vi cosθ O = (vi sin θ)t Use total time! If no time is given use: ∆x = vi2sin2θ g Jan 312:21 PM 3). maximum height: ∆y = viyt = 1/2ayt2 = (visinθ)(t) + 1/2(-9.8m/s2)(t)2 Use time at max height! (tup) Jan 312:31 PM 9 projectile.notebook February 01 and 2, 2010 vi θ viy vix 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/s2) 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 312:39 PM Example: A golf ball is hit and leaves the tee with a velocity of 25.0 m/s at 35.0 o with respect to the horizontal. What is the horizontal displacement of the ball? Find t first! s m/ .0 25 35.0 o vx vy vy θ = 35.0 x = vx t v = 25m/s x=? = v cosθ = (25 m/s)(cos 35.0) = 20.5 m/s = (25 m/s)(sin 35.0) = 14.3 m/s x = vxt =(20.5m/s)(2.9s) = vyT + 1/2gT2 = 59.45 m 2 2 0 = (14.3 m/s)T + 1/2(-9.8 m/s )T T T 0 = 14.3m/s - 4.9m/s2T T = 2.9s Jan 313:00 PM 10 projectile.notebook February 01 and 2, 2010 Practice; Problems for projectiles launched at an angle Solve Workbook Problems E pp. 25 - 27 1). vi = 45.8 m/s 2). v = 68.2 m/s 3). vi = 16.6 m/s 4 a) vi = 20.8 m/s b). the brick's max height is 11.0 m c). 22.1 m 5). ymax = 4.07 m 6). θ = 60o 7). x = 6.73 m 8). the flea's max height is 3.2 cm Jan 313:43 PM 11