Projectile Motion

Hi Dear Classmates,

                                      today Ms.Kozoriz returned our chapter 6 REINFORCEMENT. After that Ms.Kozoriz solved the problems 1-4 from chapter 6 REVIEW the solution is:

1. horizontal velocity vh=800m/s

horizontal distance dh=180m

time t=dh/vh

t=180m/800m/s

t=0.225s

vertical distance dv=1/2g.tsquare

dv=1/2.9.8m/secsquare.0.225s.0.225s

dv=0.25m

2.dv=8.0m

dh=10.0m

t=under root 2dv/g

t=under root 2.8.0m/9.8m/s

t=1.27s

vh=dh/t

vh=10m/1.27s

vh=7.8m/s

3.

    a.    vh=45m/s

theta=45 degrees

vh=cos45 degree.45m/s=31.8m/s

vv=sin45 degrees.45m/s=31.8m/s

t=2vv/g

t=2.31.8m/s/9.8m/secsquare

t=6.48s

   b.  dh=vh.t

dh=31.8m/s.6.48s

dh=206.06m

4.

    a.  v=8m/s

theta=30 degrees

vh=cos30 degrees.8m/s=6.928m/s

vv=sin30 degrees.8m/s=4m/s

t=2vv/g

t=2.4m/s/9.8m/secsquare=0.81s

   b.  dv=v2 square-v1 square/2g

dv=o-16m/s/2.-9.8

dv=-16m/s/-19.6

dv=0.82m

    c.  dh=vh.t

dh=6.928m/s.0.81s

dh=5.61m

                   the remaining 5 problems are related to our next topics so we'll do them in future.

We also solved the green book's 2-7 problems. The solutions are:

2.vh=8m/s

dv=78.4m

t=under root 2dv/g

t=4s

dh=vh.t

dh=8.4

dh=32m

3.

   a.    t=under root vv/g

t=under root 1.225/9.8

t=under root .25

t=.5 s

    b.  vh=dh/t

vh=0.40/.5

vh=0.800m/s

4. vh=2.8m/s

t=2.6s

dv=?

dh=vh.t

dh=2.8.2.6

dh=7.26m

dv=1/2gtsquare

dv=33.1m

5.

   a.  dv=1001m

vh=125km/hr=34.7m/s

t=under root 2 dv/g

t=14.3s

   b.dh=vh.t

dh=34.7.14.3

dh=496.21m

6.dv=61m

dh=23m

vh=3m/s

t=under root 2dv/g

t=3.52s

vh=dh/t

vh=6.51

7. vy=vyt+1/2gt square  since vy=o

t=under root 2y/g

t=under root 2.-0.32/-9.8

t=0.26

x=vxt=12.4m/s . 0.26s

=3.2m

          Then we revised the formulas for projectile quiz. We did quiz about projectile. Thats all we did today and TOMORROW WE'LL HAVE A TEST ON MOMENTUM and Projectile Motion so be prepared.

      

                        BEST OF LUCK

Hello everyone,

                        Today we started with chapter 6 study guide which is about motion in two dimension. The answers are 

VOCABULARY REVIEW

 1. centripetal force

 2. projectile

 3. uniform circular motion

 4. centripetal acceleration

 5. trajectory

                                         PROJECTILE  MOTION

1. To an observer at position A the ball would appear moving straight up and straight down.

2. To an observer at position B the ball would appear moving in straight line.

3. To an observer at position C the ball would appear as an arc (parabola).

4. Throughout its flight,a projectile is constantly being accelerated toward the ground. Vertical vector shrinks until it reaches the top of its flight and then it increases as it approaches the ground. The horizontal velocity vector remains constant throughout its flight. Vertical vector is under the influence of gravity and horizontal vector has no net external force acting on it.

5. Neither rock lands first. Both rocks hit the ground at same time. This is occuring because both are under the same external force that is gravity(g).

6. Time t = ?

     d =8500m

      g=9.8m/s

      t=  2d/g

      t=2.8500m/9.8m/s

      t=1734.69

      t=41.64s      answer 

7. Vertical velocity v=?

     a=vf-vi/t

     v=a.t

     v=9.8m/s(42s)

     v=410m/s

8. dh=?

    Horizontal velocity v=483km/hr=134m/s 

    d=v.t

    d=483m/s.42s

    d=5600m

       Ms.Kzoriz wrote the answers for transparency sheet 6-1 and 2 on whiteboard. The answers for 6-1 are:

1. The magnitude of the ball's velocity vector is smallest at its maximum height.

2. Equal in magnitude but opposite in direction.

3. Equal in magnitude and in direction.

4. The line for horizontal position is a straight line and for vertical position its an arc.

        The answers for 6-2 are:

1. Same vertical vector at each point.

2. Both balls fall at same acceleration.

3. Horizontal motion of red ball does not affect its vertical motion.

4. Vertical interval is greater near the bottom of diagram. Since the time interval between each two picture is same, vertical vector of ball must have increased.

5. Horizontal interval between each two pictures of ball are identical. Since the intervals represent identical time periods. Horizontal vectors of the ball must be constant.

            Ms.Kozoriz talked about upwardly launched projectiles and we started to solve the problems on page 11 PROJECTILE MOTION 2. While doing problems we enjoyed a song Harry the Human Cannonball. Thats all we did in class today. One other important thing is ourshort  momentum test will be on Tuesday not on Monday.

      Its my first blog so if you guys found any mistakes sorry about that.

Correction to yesterday's post

Horizontal velocity is independent of any force acting in the x direction (not time like you stated in your post). Vertical velocity is dependent on the acceleration due to gravity acting on the object.

Physics12Fall2010: Scribe List

Projectile motion,this is a non straight motion under gravity (g).

the horizontal component of velocity is constant i.e its independent of time,but the vertical component of the velocity (Vy)depends on time. it's 0 at the peak

october 12 scribe

October 12 2010

hi! today, we went over the Transparency worksheet.

Then miss K review a little about momentum (elastic, explosion, inelastic).

After that we have a lab with money :) and hand in after class.

Sorry!!!!!!! i don't know how to put answers or picture up. If you needed you can ask miss K

Oh yeah! we picked up study guide "Motion in Two Dimensions" and a graph. Take a look we'll go over it tomorrow...