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