Need help with Calculus and Vectors Rates of Change.
The time "t", in seconds, taken by an object dropped from a height of "s" metres to reach the ground is given by the formula (t= sqrt s/5). Determine the rate of change in time with respect to height when the object is 125m above the ground.

Question

Need help with Calculus and Vectors Rates of Change.
The time "t", in seconds, taken by an object dropped from a height of "s" metres to reach the ground is given by the formula (t= sqrt s/5). Determine the rate of change in time with respect to height when the object is 125m above the ground.

anonymous · Answer

$$t(125)=\sqrt{\frac{ 125 }{ 5 }}$$
$$t(125)=5$$

anonymous · Answer

Then I have to use this formula 
$$t(s)=  \frac{ t(s+h )- t(s)}{ h }$$
lim
h-->0

anonymous · Answer

lim h-->0 $$=\frac{ t(125+h)-t(125) }{ 5 }$$
$$=\frac{ \sqrt{\frac{ 125+h }{ 5 }} -5}{ h }$$

anonymous · Answer

idk what to do next

dumbcow · Answer

multiply/divide by conjugate 
$$\frac{\sqrt{\frac{125+h}{5}} -5}{h} *\frac{\sqrt{\frac{125+h}{5}} +5}{\sqrt{\frac{125+h}{5}} +5} = \frac{\frac{125+h}{5} -25}{h(\sqrt{\frac{125+h}{5}} +5)}$$

anonymous · Answer

I got the answer thanks anyways tho

dumbcow · Answer

ok :)