Answer:
6m/s
Explanation:
We are given that height of rock from the ground=9 m
Constant speed=10 m/s
Distance of person from rock=12 m
We have to find the rate at which distance between the rock and the second person increasing just as the rock hits the ground.
In triangle ABC
Let AB=x, BC=y,AC=z
[tex]\frac{dx}{dt}=10 m/s,\frac{dy}{dt}=0[/tex]
[tex]AB^2+BC^2=AC^2[/tex] (using pythagorous theorem )
[tex]x^2+y^2=z^2[/tex]
[tex](9)^2+(12)^2=z^2[/tex]
[tex]81+144=z^2[/tex]
[tex]225=z^2[/tex]
[tex]z=\sqrt{225}=15 m[/tex]
Differentiate w.r.t
[tex]2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}[/tex]
[tex]2(9)(10)+0=2(15)\frac{dz}{dt}[/tex]
[tex]\frac{180}{30}=\frac{dz}{dt}[/tex]
[tex]\frac{dz}{dt}=6m/s[/tex]
Hence, the rate at which distance between rock and person increasing just as the rock hits the ground=6 m/s
