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PROJECTILES 3 Types: Vertical, Horizontal, Inclined An object is launched with an initial velocity in the vertical direction. v = 0 m/s At the peak of the object’s flight, v = 0 m/s. Vertical Once the object returns to the same height it was released from, its velocity will be the same as its initial. vi Object starts out with an initial velocity in the horizontal direction. Acceleration due to gravity gives the object a velocity in the vertical direction. Horizontal An object is launched with a velocity that has vertical and horizontal components. These components can be resolved by setting up a right triangle. vi Inclined Independence of Motion The vertical and horizontal motion (free fall) does not affect the horizontal motion (constant velocity) The Horizontal Projectile vi In order to find the vertical distance travelled, we have to separate the horizontal motion from the vertical motion. dy Vertical Motion (y) viy = 0 m/s ay = -9.8 m/s2 dy = height to find the time, use: dy = viyt + ½ayt2 dy = ½ayt2 Horizontal Motion (x) vix = vi ax = 0 m/s2 dx = vixt + ½axt2 dx = vixt Note: - ax = 0!!!! Gravity affects only vertical motion. - time of flight is the same regardless of vi, a dropped bullet and a fired bullet hit the ground at the same time. Practice Problem 1 A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location. Practice Problem 2 A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball. TRAJECTORY – THE PATH OF A PROJECTILE For Horizontal projectile motion: -ax = 0 m/s2! Gravity affects only vertical motion -time of flight is the same regardless of v i, a dropped bullet and a fired bullet hit the ground at the same time. Horizontal Motion Mythbusters Bullet Drop THE INCLINED PROJECTILE Vy = 0 m/s at top viy vi θ vix Initial velocity must be resolved into components (viy and vix) HOW DO WE SOLVE THIS?? Vertical Motion (y) viy = visinθ very important!! ay = -9.81 m/s2 vfy = 0 m/s at top to find time: use vfy = viy + ayt to get time to top, then double it. to find max height: vfy2 = viy2 + 2aydy Horizontal Motion (x) Vix = vicosθ ax = 0 m/s2 t obtained from vertical to find range: dx = vixt + ½ axt2 0 PRACTICE PROBLEM #1 An airplane traveling 1001 m above the ocean at 125 km/h is going to drop a box of supplies to shipwrecked victims below. a. How many seconds before the plane is directly overhead should the box be dropped? b. What is the horizontal distance between the plane and the victims when the box is dropped? PRACTICE ANSWER #1 Vertical viy = 0 m/s ay = -9.8 m/s2 dy = 1001 m t=? d = ½ at2 1001 m = ½ (9.8 m/s2)t2 t = 14.3 s Horizontal vix = 125 km/h dx = ? t=? ax = 0 m/s2 dx = vixt dx = (125 km/h)(14.3 s) dx = (0.03472 km/s)(14.3 s) dx = 0.4965 km = 496.5 m PRACTICE PROBLEM #2 Herman the human cannonball is launched from level ground at an angle of 30° above the horizontal with an initial velocity of 26 m/s. How far does Herman travel horizontally before reuniting with the ground? PRACTICE ANSWER #2 Vertical viy = 26(sin30) ay = -9.8 m/s2 vf = 0 m/s t=? vf = viy + at 0 = 26(sin30) + (-9.8 m/s2)t -13 = -9.8 m/s2t t = 1.33 s Horizontal vix = 26(cos30) ax = 0 m/s2 dx = ? t=? dx = vixt dx = 26(cos30)(2.66 s) dx = 60.0 m EFFECT OF LAUNCH ANGLE 1. Complementary angles produce the same range. 2. Maximum range is at 45 degrees launch angle. 3. Higher angle leads to longer “hang time” 75 ° 60 ° 30 ° 15° Launch 45 ° TRAJECTORY vy = 0 vx vy vx vx viy vy vx vix vy