- #1
rea
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Ok, after some work, it seem that I am not able to do this exercises, over even get something that describe them...
I don't know even how to handle it... I can see what is asked in some way, but don't know what to plug.The next one, I'm more near, but unable to continue.
here is what I have so far:
Deducing...
[tex]a_y = -g [/tex]
[tex]v_y = -gt+V_0 [/tex]
[tex]y = -\frac{1}{2}gt^2+v_0t [/tex]
If [tex]V_y [/tex] eq 0 then the particle is at [tex]y_{max} [/tex] and [tex]t_{max} [/tex] is the time for get there...
[tex]0 = -gt_{max} + v_0 [/tex]
[tex]t = \frac{v_0}{g} [/tex]
then
[tex]y_{max} = -\frac{1}{2}g\left(\frac{v_0}{g}\right)^2+\frac{v_0v_0}{g} [/tex]
[tex] = \frac{v_0^2}{2g} [/tex]
[/tex]
Now follow a new problem, where d = h + [tex]y_{max}[/tex]
With start conditions...
[tex]
a = g [/tex]
[tex]v_0=0 [/tex]
[tex]t_0=0 [/tex]
[tex]p_0 = 0 [/tex]
And in the other end of the line:
[tex]a=g [/tex]
[tex]v_f = ? [/tex]
[tex]t_f = ? [/tex]
[tex]p_f = d = h + y_{max} [/tex]I know that this is easy, but now what?
If a projectile explode at and height of h the movement was vertical (the highest point in the vertical direction is h). The pieces fall in all directions, but all of them have the same speed v. The resistance of the iar is despreciable. Calculate the most short angle (¿acute?) that the velocity will form with the ground.
I don't know even how to handle it... I can see what is asked in some way, but don't know what to plug.The next one, I'm more near, but unable to continue.
From a height y=h a particle is launched vertically, what is the time and velocity for the particle to do contact with y = 0.
here is what I have so far:
Deducing...
[tex]a_y = -g [/tex]
[tex]v_y = -gt+V_0 [/tex]
[tex]y = -\frac{1}{2}gt^2+v_0t [/tex]
If [tex]V_y [/tex] eq 0 then the particle is at [tex]y_{max} [/tex] and [tex]t_{max} [/tex] is the time for get there...
[tex]0 = -gt_{max} + v_0 [/tex]
[tex]t = \frac{v_0}{g} [/tex]
then
[tex]y_{max} = -\frac{1}{2}g\left(\frac{v_0}{g}\right)^2+\frac{v_0v_0}{g} [/tex]
[tex] = \frac{v_0^2}{2g} [/tex]
[/tex]
Now follow a new problem, where d = h + [tex]y_{max}[/tex]
With start conditions...
[tex]
a = g [/tex]
[tex]v_0=0 [/tex]
[tex]t_0=0 [/tex]
[tex]p_0 = 0 [/tex]
And in the other end of the line:
[tex]a=g [/tex]
[tex]v_f = ? [/tex]
[tex]t_f = ? [/tex]
[tex]p_f = d = h + y_{max} [/tex]I know that this is easy, but now what?
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