Quote:
|
Originally Posted by threewingedfury
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^2-vo^2/2a and figured it reached 5.1 m, is this correct?
|
You are correct it is 40 meters, but for the wrong reason, for a 2d projectile problem, you need to break it up into x and y component.
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.