Projectile Motion

Background Information Projectile motion problems must be solved by breaking the movement of the projectile into its x (horizontal) and y (vertical) components Projectiles are any objects that are thrown or launched into the air and are subject to gravity (g = -9.8 m/s2)

Projectiles (neglecting air resistance) follow a path called a parabola The red arrows indicate movement in the y direction and blue arrows indicate movement in x direction y x y x y x y x x y

Today we are studying projectiles launched with an initial horizontal velocity from an elevated position Equations used to solve the problems are our Acceleration equations: a = ∆v/∆t xf = xi +vit + ½at2 vf2 = vi2 + 2a∆x

When solving the problems, you must solve in terms of the x (horizontal) information and separately for the y (vertical) information

HINTS… 1. When calculating the x component, we neglect all air resistance. Therefore, acceleration in the x direction is a = 0 m/s2 and this is true for ALL problems

HINTS… 2. When calculating the y component of acceleration we assume a = -9.8 m/s2 since the only force acting on the object is gravity. Also, since there is no initial velocity in the y direction, vi = 0 m/s (in y direction only)

Example A billiards ball falls off a .60 m high table with an initial velocity of 2.4 m/s. Calculate the following: a) the time required for the ball to fall to the ground b) the horizontal distance between the table’s edge and the ball’s landing position.

X Y xf = ? yf = -.60 m vi = 2.4 m/s vi = 0 m/s a = 0 m/s2 a = -9.8 m/s2 t = ? Step 1: List known and unknown variables in columns labeled X and Y yf is is negative to show that the displacement was downward…aka a negative direction

Step 2: Solve for the first unknown using the column that contains the most information Hint: Usually you need to solve for time first To solve for t: yf = yi +vit + ½at2 -.6 = 0 + 0t + ½(-9.8)t2 -.6 = -4.9t2 -4.9 -4.9 .123 = t2 .35 s = t

Step 3: Using the time, solve for any other unknowns To solve for the horizontal distance: xf = xi +vit + ½at2 xf = 0 + 2.4(.35) + .5(0)(.35)2 xf = .84 meters

Let’s Do One Together! A soccer ball is kicked horizontally off a 22 m high hill and lands 35 m away from the edge of the hill. Determine the initial horizontal velocity of the ball.

Answer t = 2.11 s vi = 16.6 m/s