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Projectile Motion Objects in projectile motion follow a parabolic path called a trajectory. If released at the same instant, a bullet shot from a gun will hit the ground at the same time as a bullet dropped from the same height. The horizontal and vertical motions of a projectile are independent, meaning they do not affect each other. Vertical motion Horizontal motion The 2-D motion of a projectile can be separated into two 1-D motions: horizontal and vertical. HPM: Horizontal Projectile Motion The horizontal motion of a projectile is always constant, if we neglect air resistance. For projectiles shot at 0°, all of the initial velocity is in the x direction. Thus, Vyi = 0 m/s. For projectiles shot at 0°, vertical displacement and velocity will always be negative. To hit the target, when should you release the package? Rules for Projectile Motion • Treat horizontal and vertical as two separate sides of the problems • TIME is the key, and the only variable that can be used for both horizontal and vertical • Horizontal Motion is always constant • vx is constant • ax = 0 m/s2 • Objects follow a parabolic shape Horizontal Projectile Motion • All of the initial velocity is in the x direction, Vyi = 0 m/s • Vertical displacement and velocity will always be negative Example problem APM: Angled Projectile Motion For projectiles shot at an angle, initial velocity is both vertical and horizontal. Horizontal velocity and initial vertical velocity can be found using trig functions. Vi θ Vyi Vx = Vi Cos θ Vyi = Vi Sin θ Vx At the end of the problem, you can recombine horizontal and vertical velocity to get the total 2-D velocity. Vx Vf Vyf Vx θ Vf Vf2 = Vx2 + Vyf2 Vyf θ = tan-1(Vyf / Vx) For APM, vy at any height is the same while going up and coming down except for direction. Velocities of projectile motion Note: Vy = 0 at the highest point. With air resistance, the actual path is shorter. Angled Projectile Motion • Initial velocity is both vertical and horizontal • Use trig functions to find vyi and vx • Vx = Vi Cos θ • Vyi = Vi Sin θ • Remember clues • vy at the top is 0 m/s • vy at any height is the same while going up and coming down except for direction Example Problem Happy Gilmore hits his shot at 55.0 m/s with an angle of 50.0° to the ground. How far did the ball travel before it lands? •vi = 55.0 m/s •θ = 50.0° •ay = -9.81 m/s2 •∆x = ? vi = 55.0 m/s θ = 50.0° • Find vx and vyi vx = vi Cos θ = 55.0 m/s Cos (50.0°) = 35.4 m/s Sin θ = 55.0 m/s Sin (50.0°) = 42.1 m/s vyi = vi ay = -9.81 m/s2 ∆x = ? vi = 55.0 m/s θ = 50.0° ay = -9.81 m/s2 vyi = 42.1 m/s ∆x = ? vx = 35.4 m/s • What do we need to find ∆x? • Time! Find time from the vertical side • ∆y = vyi ∆t + ½ ay ∆t2 0 m = (42.1 m/s) ∆t + ½ (-9.81 m/s2) ∆t2 - (42.1 m/s) ∆t = ½ (-9.81 m/s2) ∆t2 (42.1 m/s) = ½ (9.81 m/s2) ∆t ∆t = 8.58 s vi = 55.0 m/s θ = 50.0° ∆t = 8.58 s ay = -9.81 m/s2 vyi = 42.1 m/s • Now we can find ∆x • ∆x = vx ∆t = (35.4 m/s)(8.58 s) = 304 m ∆x = ? vx = 35.4 m/s Where Should We Aim the Cannon? At or Above the monkey? Above the Monkey At the Monkey Explanation of monkey experiment Ranges of projectiles versus angle. Another interesting application If speed is great enough… That’s how the space shuttle and satellites orbit the earth.