Help with projectile motion

  • Thread starter ballahboy
  • Start date
  • #1
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hi im new to the board..
my class recently did a lab using water rockets as projections. Our rocket went 30.2m in 1.73 seconds. can u guys help me find Vo, V, angle of launch and angle of impact? i need help on this a lot. can u guys also show me the steps in doin this.
thanks
 

Answers and Replies

  • #2
1,015
1
30.2 m in th x direction?
 
  • #3
1,015
1
If so

this may work
I am tired so please forgive me if I error

[tex] 30.2=Vo*cos(\theta)t [/tex]

[tex] let A = Vo*cos(\theta)=\frac{30.2}{1.73}[/tex]

[tex] 0=Vo sin (\theta) - g\frac{t}{2} [/tex]

[tex] let B = g\frac{t}{2}=Vo sin (\theta)[/tex]

[tex] Vo=\sqrt{A^2+B^2} = \sqrt{ (\frac{30.2}{1.73})^2+(g\frac{t}{2})^2) }
[/tex]

[tex] Vo = apprx: 19.406 [/tex]Keep exact value call C

[tex]30.2=Vo*cos(\theta)t[/tex]
[tex] 30.2=C*cos(\theta)1.73[/tex]

[tex] \frac{30.2}{C*1.73}=cos(\theta)[/tex]

[tex] cos^{-1}(\frac{30.2}{C*1.73})=(\theta)=approx 25.9 degrees[/tex]= angle of lanuch

angle of impact should be 154.0 from outside
impact should simply be

[tex] V=Vo+at [/tex]

take it from when projectile is at max height onward

[tex] V=0-g\frac{t}{2}[/tex]
[tex] V=-8.477 m/s [/tex]
 
Last edited:
  • #4
1,015
1
ballahboy said:
hi im new to the board..
my class recently did a lab using water rockets as projections. Our rocket went 30.2m in 1.73 seconds. can u guys help me find Vo, V, angle of launch and angle of impact? i need help on this a lot. can u guys also show me the steps in doin this.
thanks

Summary

Vo, V, angle of launch and angle of impact

Vo: approx 19.41 m/s
V= approx -8.477 m/s
Angle of Launch approx 25.9 degrees
Angle of Impact approx 154.1 degrees
 
  • #5
34
0
wow thank u so much
 
  • #6
34
0
sorry im still confused, how did u get this 0=(Vo sin theta)-g*t/2

also y is Vo=0 at the bottom but we proved it was 19.41?
 
Last edited:
  • #7
1,015
1
ballahboy said:
sorry im still confused, how did u get this 0=(Vo sin theta)-g*t/2

also y is Vo=0 at the bottom but we proved it was 19.41?

We divide the projectile motion in half and only examine the last half of the trip... since it is a smooth parapbolic path, the time it takes to go from the top to the bottom will be half the total time.

When you examine it at the top peak of its height to the bottom, at the peak the Vo will be zero because there is no velecity.

so then to get V you would have the velocity of impact you would get V=Vo-gt... but t in this instance is the total t in half. so V=0-gt/2


-------------------------------------
for
[tex] 0=Vo sin (\theta) - g\frac{t}{2} [/tex]

I did the same dividing stat, I just decided to examine the first half of the trip.
We know the final velocity will be 0 at the top so I just solved it that way... again t would be in half this time.
 

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