Respuesta :
Answer:
Rocks will pass each other at 20.4m.
Explanation:
Alice drops the rock from [tex]y_{a,i}=40m[/tex] (y is in the vertical axis) at [tex]v_{a,i}=0[/tex].
Bob throws the rock from [tex]y_{b,i}=0m[/tex] at [tex]v_{b,i}}=20\frac{m}{s}[/tex].
[tex]g[/tex] is gravity's acceleration.
Position equation for alice's rock:
[tex]y_{a}(t)=y_{a,i} +v_{a,i}t - \frac{1}{2}gt^{2} =40m-\frac{1}{2}gt^{2}[/tex]
Position equation for Bob's rock:
[tex]y_{b}(t)=y_{b,i} + v_{b,i}t - \frac{1}{2}gt^{2} =20\frac{m}{s}t -\frac{1}{2}gt^{2}[/tex]
We we'll first find the time [tex]t_{0}[/tex] at which the rocks meet. For this, we will equalize Bob's and Alice's equations:
[tex]y_{a}(t_{0})=y_{b}(t_{0})[/tex]
[tex]40m-\frac{1}{2}gt_{0}^{2}=20\frac{m}{s}t_{0} -\frac{1}{2}gt_{0}^{2}[/tex]
⇒[tex]20\frac{m}{s}t_{0}=40m[/tex] ⇒ [tex]t_{0}=2s[/tex]
So, we can take [tex]t_{0}=2[/tex] and just put it in any of the two laws of motion to see at what height the rocks meet. We will take Alice's equation (using g=9.8m/s):
[tex]y_{a}(2) =40m-\frac{1}{2}g2^{2}=20.4m[/tex]
Rocks will pass each other at 20.4m.
