Answer:
Assume that [tex]g =9.81\; \rm m\cdot s^{-1}[/tex], and that the air resistance on the stone is negligible.
a.
Height of the stone: [tex]389.5\; \rm m[/tex] (above the ground.)
Velocity of the stone: [tex]\left(-110.5\; \rm m \cdot s^{-1}\right)[/tex] (the stone is travelling downwards.)
b.
Height of the stone: [tex]629.5\; \rm m[/tex] (above the ground.)
Velocity of the stone: [tex]\left(-86.5\; \rm m \cdot s^{-1}\right)[/tex] (the stone is travelling downwards.)
Explanation:
If air resistance on the stone is negligible, the stone would be accelerating downwards at a constant [tex]a = -g = -9.81\; \rm m \cdot s^{-2}[/tex].
Let [tex]h_0[/tex] denote the initial height of the stone (height of the stone at [tex]t = 0[/tex].)
Similarly, let [tex]v_0[/tex] denote the initial velocity of the stone.
Before the stone reaches the ground, the height [tex]h[/tex] (in meters) of the stone at time [tex]t[/tex] (in seconds) would be:
[tex]\displaystyle h(t) = -\frac{1}{2}\, g \cdot t^{2} + v_0 \cdot t + h_0[/tex].
Similarly, before the stone reaches the ground, the velocity [tex]v[/tex] (in meters-per-second) of the stone at time [tex]t[/tex] (in seconds) would be:
[tex]v(t) = -g\cdot t + v_0[/tex].
In section a., [tex]h_0 = 1000\; \rm m[/tex] while [tex]v_0 = -12\; \rm m\cdot s^{-1}[/tex] (the stone is initially travelling downwards.) Evaluate both [tex]h(t)[/tex] and [tex]v(t)[/tex] for [tex]t = 10\; \rm m \cdot s^{-1}[/tex]:
[tex]\begin{aligned} h(t) &= -\frac{1}{2}\, g \cdot t^{2} + v_0 \cdot t + h_0 \\ &= -\frac{1}{2}\ \times 9.81\; \rm m\cdot s^{-2}\times (10\; \rm s)^{2} \\&\quad\quad + \left(-12\; \rm m \cdot s^{-1}\right) \times 10\; \rm s + 1000\; \rm m \\[0.5em] &= 389.5\; \rm m \end{aligned}[/tex].
Indeed, the value of [tex]h(t)[/tex] at [tex]t = 10\; \rm m \cdot s^{-1}[/tex] is greater than zero. The stone hasn't yet hit the ground, and both the representation for the height of the stone and that for the velocity of the stone are valid.
[tex]\begin{aligned} v(t) &= -g\cdot t + v_0 \\ &= -9.81\; \rm m\cdot s^{-2}\times 10\; \rm s - 12\; \rm m\cdot s^{-1} \\ &= -110.5\; \rm m \cdot s^{-1} \end{aligned}[/tex].
The value of [tex]v(t)[/tex] at [tex]t = 10\; \rm m \cdot s^{-1}[/tex] is negative, meaning that the stone would be travelling downwards at that time.
In section b., [tex]h_0 = 1000\; \rm m[/tex] while [tex]v_0 = 12\; \rm m\cdot s^{-1}[/tex] (the stone is initially travelling upwards.) Evaluate both [tex]h(t)[/tex] and [tex]v(t)[/tex] for [tex]t = 10\; \rm m \cdot s^{-1}[/tex]:
[tex]\begin{aligned} h(t) &= -\frac{1}{2}\, g \cdot t^{2} + v_0 \cdot t + h_0 \\ &= -\frac{1}{2}\ \times 9.81\; \rm m\cdot s^{-2}\times (10\; \rm s)^{2} \\&\quad\quad + 12\; \rm m \cdot s^{-1} \times 10\; \rm s + 1000\; \rm m \\[0.5em] &= 629.5\; \rm m \end{aligned}[/tex].
Verify that the value of [tex]h(t)[/tex] at [tex]t = 10\; \rm m \cdot s^{-1}[/tex] is indeed greater than zero.
[tex]\begin{aligned} v(t) &= -g\cdot t + v_0 \\ &= -9.81\; \rm m\cdot s^{-2}\times 10\; \rm s + 12\; \rm m\cdot s^{-1} \\ &= -86.5\; \rm m \cdot s^{-1} \end{aligned}[/tex].
Similarly, the value of [tex]v(t)[/tex] at [tex]t = 10\; \rm m \cdot s^{-1}[/tex] is negative because the stone would be travelling downwards at that time.
