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## Projectiles

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**Definition**• 2 dimensional motion of an object after it has been thrown, shot, etc • An object that has only the force of gravity working on it.**an object dropped from rest**• an object thrown vertically upwards • an object thrown horizontally • an object thrown upwards at an angle**Common Misconception**• If an object is moving upwards, there must be a force acting in the upward direction • Recall: a force is required to continue acceleration**Consider:**• cannonball fired in a gravity free environment • No force needed to sustain the motion of the cannonball**What happens when we compare the motion of the cannonball in**gravity free and gravity rich environments?**Gravity is the only force acting on the object, causing**downward acceleration (9.81 m/s2) • Parabolicfall of a projectile is the result of horizontal (x) and vertical (y) motion occurring at the same time**Problem solving**• Assume: air resistance is negligible (it doesn’t affect the object) • We study the motion as separate vertical and horizontal motions.**More Problem Solving Tips**• Time is dependent on Vertical Motion • Use vertical motion formulas to determine • Range (horizontal distance, x) is dependent on time**Problem solving (continued)**Vertical component (y- direction) • downward acceleration of 9.81 m/s2 • Which means that we can substitute g into any equation that has acceleration • at highest point in trajectory Vy = 0 m/s • Vy = Vy0 + gt • y = y0 + Vy0 t + ½ g t2 • Where y is the vertical distance • (in most problems y0 will be zero because the object lands at the same height as it was launched) • Vy2 = Vy02 + 2gy**Problem solving (continued)**Horizontal component (x- direction) • no acceleration Vx is constant • x = x0 + Vx0 t**Homework**Read pages 147 – 149 Answer questions 1-3 on page 150**Projectiles Launched Horizontally**• http://www.stmary.ws/highschool/physics/home/animations3/horizProjectileDropped.html - Vy0 (initial vertical velocity) = 0 m/s • To find time: y = ½ gt2 • To find range: x = Vx t**Practice Problem**A stunt driver on a motorcycle speeds horizontally off a 50.0 m cliff. a. How long is the motorcycle in the air?**How fast must the motorcycle leave the cliff top if it is to**land on the safety mats on level ground below, 90.0 m from the base of the cliff?**A ball is thrown at 29.3 m/s from a cliff 43.9 m above the**ground. (Neglect air resistance) How long does the ball take to hit the ground? How far from the base of the cliff does the ball land?**Projectiles Launched at an Angle**• Projectilehas an initial vertical velocity (Viy) as well as a horizontal velocity (Vix) Viy = Vi sin Q Vix = VicosQ**Vertical Motion of Projectile Launched at an angle**• Vertical velocity acts like an object tossed straight up in the air • Vi decreases to zero, then increases in the negative direction • Constant acceleration downward • Total Time of Flight: t = -2 Viy g**Vertical Motion of Projectile Launched at an angle**• Maximum height (ymax) Vy = 0 m/s ymax = Viy2 / (2g)**Horizontal Motion**• Constant velocity • Range (horizontal distance, X) x = Vix t