Monday, October 18, 2010

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





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