Hi,
I have read the Mechanics (review).
The question:
A person jumps from a rock at 3.6m altitude. The initial horizontal velocity is 1m/sec and the vertical 3m/sec.
What are the position coordinates at the top of the jump?
What is the velocity when he lands into the water and what are the landing coordinates?
Is this a symetrical projectile or a asymetrical projectile?
Louis
You may obviously choose your initial coordinates. Just be consistent: If your 0,0 is the rock that your example leaps from, it must remain as the origin for the remainder of the maths to be meaningful.
Assume the vertical component is 3 m/s in the upward direction, so in solving the first part of the equation, you have two answers: the 'top-of-jump' position, and the new coordinates for the second part of the solution you'll need to do, in the same way.
You can then graph the entire trajectory.
As far as symmetrical v. asymmetrical projectiles go, that sounds sort of vague to me, and unanswerable as such.
I imagine that given the variables and constraints of the problem that your teacher isn't asking about moments of inertia or rotation of the body...
If you meant to say trajectory instead of projectile it would make more sense to me.
Assume the vertical component is 3 m/s in the upward direction, so in solving the first part of the equation, you have two answers: the 'top-of-jump' position, and the new coordinates for the second part of the solution you'll need to do, in the same way.
You can then graph the entire trajectory.
As far as symmetrical v. asymmetrical projectiles go, that sounds sort of vague to me, and unanswerable as such.
I imagine that given the variables and constraints of the problem that your teacher isn't asking about moments of inertia or rotation of the body...If you meant to say trajectory instead of projectile it would make more sense to me.
Thank you for help,
so first I need the time. I get it by Vy=Voy+gt V=3 - 9.81 t ; -3=-9.81 t ; -3 divided by -9.81 = 0.306s ;
For displacement: X = Xo + Vox t ; X = 0+1 0.306 ; X = 0.31m
Y = 3.6 + 3 0.306 + 0.5 9.81 0.306^2
Y = 4.06m
I think the top of the jump must be half of the time so 4.06m / 2 = 2.03m
3.6m+2.03m = 5.63m for Y and 0.31m /2 = 0.15m for X ;
Velocity at landing is Vy = Voy+g t so Vy = 3+9.81 0.306
Vy = 6 m/s
That are my solutions. I do not be wrong, do I?
Louis
so first I need the time. I get it by Vy=Voy+gt V=3 - 9.81 t ; -3=-9.81 t ; -3 divided by -9.81 = 0.306s ;
For displacement: X = Xo + Vox t ; X = 0+1 0.306 ; X = 0.31m
Y = 3.6 + 3 0.306 + 0.5 9.81 0.306^2
Y = 4.06m
I think the top of the jump must be half of the time so 4.06m / 2 = 2.03m
3.6m+2.03m = 5.63m for Y and 0.31m /2 = 0.15m for X ;
Velocity at landing is Vy = Voy+g t so Vy = 3+9.81 0.306
Vy = 6 m/s
That are my solutions. I do not be wrong, do I?
Louis