Answer to Question #124009 in Electrical Engineering for Darren Regondola

Question

If another stone is thrown vertically upward after 1 second at twice the initial velocity of the first, how far above the ground will the stones be at the same level?

Answer 1

one-dimensional equation of motion due to gravity:

$$h(t)=h_0+v_0t-\frac{1}{2}g t^2$$

$h_0=0$ - the ground

$v_0$ - the initial velocity

$g=9.8m/s$ - the acceleration due to gravity

then the equation for the first stone:

$$h_1(t)=v_0t-\frac{1}{2}g t^2$$

If another stone is thrown vertically upward after 1 second at twice the initial velocity of the first, then the equation for the second stone:

$$h_2(t)=2v_0(t-1)-\frac{1}{2}g (t-1)^2$$

we find the moment in time corresponding to the same level from the equation:

$$h_1(t)=h_2(t)$$ $$v_0t-\frac{1}{2}g t^2=2v_0(t-1)-\frac{1}{2}g (t-1)^2$$ $$v_0t-2v_0t-gt=-2v_0-\frac{1}{2}g$$ $$t=\frac{1}{2}\frac{4v_0+g}{v_0+g}$$

now we substitute this solution into the equation of motion and find the height value:

$$h=v_0t-\frac{1}{2}gt^2=v_0\frac{1}{2}\frac{4v_0+g}{v_0+g}-\frac{1}{2}g(\frac{1}{2}\frac{4v_0+g}{v_0+g})^2=\frac{1}{8}\frac{16v_0^3+4v_0^2g-4v_0g^2-g^3}{(v_0+g)^2}$$

additionally, find initial speed requirements:

$$h=0 \rightarrow v_0=g/2=4.9m/s$$

in order for this task to have physical meaning, it is necessary that the initial speed of the first stone be more than half the acceleration value under the influence of gravity; with a minimum initial speed the first stone will move towards the second stone

consider the case when the first and second stones move in the same direction (up):

$$\begin{cases} h_1(t)=h_2(t) \\ v_1(t)\geq0 \end{cases}$$ $$\begin{cases} t=\frac{1}{2}\frac{4v_0+g}{v_0+g} \\ v_0-\frac{1}{2}gt\geq 0 \end{cases}$$ $$v_0 \geq \frac{g}{2}(\sqrt{3}+1)\approxeq13.4 m/s$$

Answer:

$level:\space h=\frac{1}{8}\frac{16v_0^3+4v_0^2g-4v_0g^2-g^3}{(v_0+g)^2},\space where\space g=9.8m/s,\space the \space initial \space velocity: \space v_0>g/2$