Kinematics - 6 - 2D Horizontal Launch Angle

3.8 2-D Motion Horizontal Launch Angle (Type 3) • YouTube - Mythbusters Dropped Bullet • When an object moves both horizontally and vertically, the motions are independent of each other. • Vertical acceleration is due to gravity: ay = -9.80 m/s2 • Horizontal acceleration is zero! No Fx, uniform motion in x-direction. • Each motion is worked out separately. Vertical Component 1 2 d y = v1 y t + a y t 2 • For horizontal launch, V1y = 0. • To find V2 we can still use: v2 y = v1 y + a y t (v ) = (v ) 2 2y 2 1y + 2a y d y Horizontal Component d x = vx t Organization! • In 2-D motion, keeping ‘x’ and ‘y’ numbers separated is very important. • Use a data table to help you. X Y ∆dx = ____ m ∆dy = ____ m Vx = _____ m/s V1y = ____ m/s V2y = ____ m/s ay = -9.80 m/s2 ∆t = ____ s ∆t = ____ s - Convert units first. - Start solving in column with fewer unknowns. Ex #1 • A ball rolls off of a flat roof with a horizontal velocity of 12.0 m/s. The roof is 15.0 m high. How far from the base of the building will the ball land? Answer 1. Find ∆t X Y ∆dx = ? ∆dy = -15.0 m Vx = 12.0 m/s V1y = 0 m/s V2y ay = -9.80 m/s2 ∆t = ? ∆t = ? 2. Find ∆dx d y = v1 y t + a y ( t ) 1 2 1 − 15 = 0 + ( −9.8)(t ) 2 2 − 15 − 4.9( t ) 2 = − 4.9 − 4.9 − 15 = t − 4.9 t = 1.74964s Δd x = v x Δt = 12  1.74964 = 20.9956 = 21.0 m 2 Ex #2 • Heather throws a ball horizontally off the top of a 35.0 m cliff with a horizontal velocity of 4.60 m/s. What is its velocity upon impact? X Y ∆dx ∆dy = -35.0 m Vx = 4.60 m/s V1y = 0 m/s V2y = ? 1. Find ∆t ay = -9.80 m/s2 ∆t = ? ∆t = ? d y = v1 y t + 12 a y ( t ) 2 1 − 35 = 0 + ( −9.8)(t ) 2 2 − 35 − 4.9( t ) 2 = − 4.9 − 4.9 − 35 = t − 4. 9 t = 2.67261s 2. Find v2y v2 y = v1 y + at = 0 + (−9.8)  2.67261 = −26.1916 m/s X Y ∆dy = -35.0 m Vx = 4.60 m/s V1y = 0 m/s V2y = -26.1916 m/s ay = -9.80 m/s2 ∆t = 2.6726 s ∆t = 2.6726 s v 2 = 4.62 + 26.19162 = 26.5925 m/s  opp    = tan   adj  −1  26.1916  = tan    4.6  = 80.0388 −1 3. Find v2 4.60 m/s θ v2 26.19 m/s Try It! • A ball rolls off of a flat roof that is 24.0m high. If the ball lands 22.0m from the base of the building, what is the initial velocity? Answer 1. Find ∆t X Y ∆dx = 22.0 m ∆dy = -24.0 m Vx = ? V1y = 0 m/s V2y ay = -9.80 m/s2 ∆t = ? ∆t = ? 2. Find ∆vx d y = v1 y t + a y ( t ) 1 2 1 − 24 = 0 + ( −9.8)(t ) 2 2 − 24 − 4.9( t ) 2 = − 4.9 − 4.9 − 24 = t − 4.9 t = 2.2131s v x = Δd x / t = 22 / 2.2131 = 9.94 m/s 2 Ex #3 From the roof of a building 50.0m high, a ball is pushed horizontally with a velocity of 5.00m/s. A) What is the ball’s position after one second? X Y ∆dx = ? ∆dy = ? m Vx = 5.00 m/s V1y = 0 m/s ay = -9.8 m/s2 ∆t = 1 s ∆t = 1 s d x = vx t 1 d y = v1 y t + a y t 2 2 d x = (5)(1) 1 d y = 0 + ( −9.8)(1) 2 2 d x = 5m (5m,45.1m ) d y = −4.9m X ∆dx = ? Vx = 5.00 m/s B) How far from the building will the ball be when it touches the ground? 1 d y = v1 y t + a y t 2 2 1 − 50 = 0 + ( −9.8)(t ) 2 Y 2 ∆dy = -50.0 m V1y = 0 m/s V2y ay = -9.80 m/s2 ∆t = ? ∆t = ? t = 3.19s d x = vx t d x = (5)(3.19) d x = 16.0m Ex #4 A ball is pushed horizontally from a roof at a velocity of 6.00m/s. It hits the ground 30.0m away from the base of the building. How high is the roof? X Y ∆dx = 30.0 m ∆dy = ? m Vx = 6.00 m/s V1y = 0 m/s V2y ay = -9.80 m/s2 ∆t = ? ∆t = ? h = 123m d x = vx t 30 = (6)(t ) t = 5.00s 1 2 d y = v1 y t + a y t 2 1 d y = (0)(5) + (−9.8)(5) 2 2 d y = −122.5m HW: Quantum pg 251 #2ab, 4, 5a, 6, 8