Respuesta :
The stone's acceleration, velocity, and position vectors at time [tex]t[/tex] are
[tex]\mathbf a(t)=-g\,\mathbf j[/tex]
[tex]\mathbf v(t)=v_{i,x}\,\mathbf i+\left(v_{i,y}-gt\right)\,\mathbf j[/tex]
[tex]\mathbf r(t)=v_{i,x}t\,\mathbf i+\left(y_i+v_{i,y}t-\dfrac g2t^2\right)\,\mathbf j[/tex]
where
[tex]g=9.80\dfrac{\rm m}{\mathrm s^2}[/tex]
[tex]v_{i,x}=\left(28.0\dfrac{\rm m}{\rm s}\right)\cos43.0^\circ\approx20.478\dfrac{\rm m}{\rm s}[/tex]
[tex]v_{i,y}=\left(28.0\dfrac{\rm m}{\rm s}\right)\sin43.0^\circ\approx19.096\dfrac{\rm m}{\rm s}[/tex]
and [tex]y_i[/tex] is the height of the building and initial height of the rock.
(a) After 6.1 s, the stone has a height of 5 m. Set the vertical component [tex](\mathbf j)[/tex] of the position vector to 5 m and solve for [tex]y_i[/tex]:
[tex]5\,\mathrm m=y_i+\left(19.096\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.1\,\mathrm s)^2[/tex]
[tex]\implies\boxed{y_i\approx70.8\,\mathrm m}[/tex]
(b) Evaluate the horizontal component [tex](\mathbf i)[/tex] of the position vector when [tex]t=6.1\,\mathrm s[/tex]:
[tex]\left(20.478\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)\approx\boxed{124.92\,\mathrm m}[/tex]
(c) The rock's velocity vector has a constant horizontal component, so that
[tex]v_{f,x}=v_{i,x}\approx20.478\dfrac{\rm m}{\rm s}[/tex]
where [tex]v_{f,x}[/tex]
For the vertical component, recall the formula,
[tex]{v_{f,y}}^2-{v_{i,y}}^2=2a\Delta y[/tex]
where [tex]v_{i,y}[/tex] and [tex]v_{f,y}[/tex] are the initial and final velocities, [tex]a[/tex] is the acceleration, and [tex]\Delta y[/tex] is the change in height.
When the rock hits the ground, it will have height [tex]y_f=0[/tex]. It's thrown from a height of [tex]y_i[/tex], so [tex]\Delta y=-y_i[/tex]. The rock is effectively in freefall, so [tex]a=-g[/tex]. Solve for [tex]v_{f,y}[/tex]:
[tex]{v_{f,y}}^2-\left(19.096\dfrac{\rm m}{\rm s}\right)^2=2(-g)(-124.92\,\mathrm m)[/tex]
[tex]\implies v_{f,y}\approx-53.039\dfrac{\rm m}{\rm s}[/tex]
(where we took the negative square root because we know that [tex]v_{f,y}[/tex] points in the downward direction)
So at the moment the rock hits the ground, its velocity vector is
[tex]\mathbf v_f=\left(20.478\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(-53.039\dfrac{\rm m}{\rm s}\right)\,\mathbf j[/tex]
which has a magnitude of
[tex]\|\mathbf v_f\|=\sqrt{\left(20.478\dfrac{\rm m}{\rm s}\right)^2+\left(-53.039\dfrac{\rm m}{\rm s}\right)^2}\approx\boxed{56.855\dfrac{\rm m}{\rm s}}[/tex]
(d) The acceleration vector stays constant throughout, so
[tex]\mathbf a(t)=\boxed{-g\,\mathbf j}[/tex]
