09182006, 01:18 PM  #1 (permalink) 
Upright

physics projectile motion and uniform circular motion
Ok I had a test today with these questions on it, and I'm completly lost
A projectile reaches its maximum height two seconds after launch. At its max height its speed is 10 m/s. How far does the particle travel? What are the x and y components? How high does it go? Is it safe to say that the particle travels 40 m because of conservation of energy, because I figured 10 m/s was the initial velocity too? Is the x component 10i and y component 0j? I did v^2vo^2/2a and figured it reached 5.1 m, is this correct? A particle is in uniform circular motion about the origin of an xy coordinate system moving clockwise with a period of 7 seconds. At one instant its position vector is 2i3j Find speed and magnitude of the acceleration. I completely dont understand this one! 
09182006, 03:42 PM  #2 (permalink) 
Go Cardinals
Location: St. Louis/Cincinnati

1) Don't confuse speed with velocity.
2) Is this a calculusbased physics course or algebrabased? Here are some of my thoughts: At t=2, the ycomponent of speed will be zero. The xcomponent of speed does NOT change during the course of the flight. You can figure out the ycomponent using the acceleration due to gravity of two seconds (2 seconds to reach the top, two seconds to fall gives you two seconds of the ycomponent of speed). For the height and distance (range) problems, you are going to need the angle at which the projectile was shot. Without the angle, you cannot solve the problem.
__________________
Brian Griffin: Ah, if my memory serves me, this is the physics department. Chris Griffin: That would explain all the gravity. Last edited by soccerchamp76; 09182006 at 04:54 PM.. Reason: Automerged Doublepost 
09182006, 05:41 PM  #3 (permalink)  
Devils Cabana Boy
Location: Central Coast CA

Quote:
Since when the projectile is at its peak, there is not velocity in the x direction, we know that the velocity in the y direction is 10 m/s. since we know that it traveled up for 2 seconds, it will travel down for 2 seconds as well, giving us 4 seconds of time in the air, moving in the x direction at 10 m/s giving us 40 meters. Now we need to find the y component: we know that after 2 seconds it is at its peak Y = ½ g*t^2 +Vyo*t +Yo. Yo = 0, initial y position is 0 Vyo = ?, the initial y velocity Solve for Vyo; 19.6 m/s So we have Vyo = 19.6 m/s and Vxo as 10 m/s So since it travels up for 2 seconds we can again use the Y function to solve for the max height now that we have the Vyo. Vyo = 19.6 meters high. I’m off to pick up dinner, I’ll try to answer b when I am eating.
__________________
Donate Blood! "Love is not finding the perfect person, but learning to see an imperfect person perfectly." Sam Keen 

09192006, 08:18 PM  #4 (permalink)  
Psycho

Quote:
Now find the circumfrence of the circle. You now have the distance the particle is travelling over a certain amount of time. So you have the speed. As for the centripetal acceleration it is given by a=(v^2)/r
__________________
"I am the wrath of God. The earth I pass will see me and tremble." Klaus Kinski as Don Lope de Aguirre 

10012006, 07:55 AM  #5 (permalink)  
Location: Waterloo, Ontario

Quote:
You can determine the maximum height because you know that the y component reaches the maximum height in 2 seconds. This is the equivalent of finding the distance an object travels after accelerating at 1G for 2 seconds... You can determine the distance of the projectile because you know the time of flight, 2 + 2 = 4 seconds, and you know it's x component velocity, 10 m/s. This works out to 40m. As you can see, the angle was not needed, here... PS. I'm sorry I got to this thread so late. This forum is too slow to check too often... 

Tags 
circular, motion, physics, projectile, uniform 
Thread Tools  

