# Two-Dimensional Motion Projectiles launched at an angle

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Two-Dimensional Motion Projectiles launched at an angle
Chapter 3 Two-Dimensional Motion Projectiles launched at an angle

Some Variations of Projectile Motion
An object may be fired horizontally The initial velocity is all in the x-direction vi = vx and vy = 0 All the general rules of projectile motion apply

Projectile Motion at an angle

How are they different? Projectiles Launched Horizontally
The initial vertical velocity is 0. The initial horizontal velocity is the total initial velocity. Projectiles Launched At An Angle Resolve the initial velocity into x and y components. The initial vertical velocity is the y component. The initial horizontal velocity is the x component.

x-direction ax = 0 vx = vix = vicosΘi = constant x = vixt This is the only equation in the x-direction since there is constant velocity in that direction Initial velocity still equals final velocity

y-direction viy = visinΘi Free fall problem a = g Object slows as it goes up (-9.8m/s2) Uniformly accelerated motion, so the motion equations all hold JUST LIKE STOMP ROCKETS Symmetrical

Problem-Solving Strategy
Resolve the initial velocity into x- and y-components Treat the horizontal and vertical motions independently Make a chart again, showing horizontal and vertical motion Choose to investigate up or down. Follow rules of kinematics equations

Solving Launched Projectile Motion
vi = (r , Θ) = (vix , viy) Horizontal Vertical a = vi = vf = t = x = vix # # = range a = vi = vf = t = x = +/- 9.8m/s2 = g 0 or viy viy or 0 # Max height = y UP or DOWN INVESTIGATION… Where do the resolved components go?

Projectile Motion at an angle

Example 1: The punter for the Steelers punts the football with a velocity of 27 m/s at an angle of 30. Find the ball’s hang time, maximum height, and distance traveled (range) when it hits the ground. (Assume the ball is kicked from ground level.) Looking for: Total time (t) Max height (y) Range (x) Given: vi = (27m/s, 30o)

What do we do with the given info?
vi = (27m/s, 30o) What are the units? vi = (23.4m/s, 13.5m/s) “resolved” vector m/s 27m/s 30o Viy = 27sin30 Viy = 13.5 Viy = 13.5m/s Vix = 27cos30 Vix = 23.4 Vix = 23.4m/s

So where does this info “fit” in the chart?
Horizontal Vertical a = vi = vf = t = x = 23.4m/s a = vi = vf = t = x = Viy if solving “up” = 13.5m/s Viy if solving “down” = 13.5m/s

Pick a “side” to solve – symmetry
Up: Horizontal Vertical a = vi = vf = t = x = 23.4m/s a = vi = vf = t = y = - 9.8m/s2 13.5m/s vf2 = vi2 + 2gy y = 9.3m vf = vi + gt t = 1.38s

Projectile Motion at an angle

Pick a “side” to solve – symmetry
Down: Horizontal Vertical a = vi = vf = t = x = 23.4m/s a = vi = vf = t = y = 9.8m/s2 13.5m/s vf2 = vi2 + 2gy y = 9.3m vf = vi + gt t = 1.38s

Projectile Motion at an angle

On the horizontal side Max height occurs midway through the flight.
We found t = 1.38s both directions (up and down). How long is the projectile in the air? DOUBLE this time for total air time t = 1.38 x 2 = 2.76s What about range? x = vt x = (23.4)(2.76) x = 64.6m = RANGE

This tells us… Now we only need an initial velocity vector to determine all of the information we need to have a detailed description of where an object is in its path. vi = (r , Θ) = (vix , viy)

Maximum Range vs. Maximum Height
What angle of a launched projectile gets the maximum height? What angle of a launched projectile gets the maximum range? 90o 45o

Projectile Motion at Various Initial Angles
Complementary values of the initial angle result in the same range The heights will be different The maximum range occurs at a projection angle of 45o

Non-Symmetrical Projectile Motion
Follow the general rules for projectile motion Break the y-direction into parts up and down symmetrical back to initial height and then the rest of the height

your homework … We are going to see what kind of job Hollywood writers and producers would do on their NECAP assessments… Watching a clip of the Bus Jump, use the timer provided to time the flight of the bus and then do the actual calculations for homework.