Question

the equation that gives the distance fallen, s, by an object in time, t, assuming that the object starts at rest is: s = 1/2 g t^2 where g is the acceleration of gravity. solve the equation for time. a t = sqrt(4s/g) b t = 2s/g c t = sqrt(g/(2s)) d t = sqrt(2s/g)

Explanation:

Step1: Isolate $t^2$

Starting with $s=\frac{1}{2}gt^{2}$, multiply both sides by $2$ to get $2s = gt^{2}$. Then divide both sides by $g$: $t^{2}=\frac{2s}{g}$.

Step2: Solve for $t$

Take the square - root of both sides. Since $t$ represents time and is non - negative in this context, $t=\sqrt{\frac{2s}{g}}$.

Answer:

D. $t = \sqrt{\frac{2s}{g}}$