Answer:
[tex]\displaystyle g = \frac{2(h-vt)}{t^2}[/tex]
Step-by-step explanation:
We are given the equation:
[tex]\displaystyle h = vt+\frac{1}{2}gt^2[/tex]
And we want to solve for g, the acceleration due to gravity.
Proceed with algebra:
[tex]\displaystyle \begin{aligned} h & = vt + \frac{1}{2}gt^2 \\ \\ h - vt & = \frac{1}{2}gt^2 \\ \\ gt^2& = 2(h-vt) \\ \\ g &= \frac{2(h-vt)}{t^2}\end{aligned}[/tex]
In conclusion:
[tex]\displaystyle g = \frac{2(h-vt)}{t^2}[/tex]
Answer:
g = [tex]\frac{2(h - vt)}{t^2}[/tex]
Step-by-step explanation:
Given
h = vt + [tex]\frac{1}{2}[/tex] gt² ( subtract vt from both sides )
h - vt = [tex]\frac{1}{2}[/tex] gt² ( multiply both sides by 2 to clear the fraction )
2(h - vt) = gt² ( divide both sides by t² )
[tex]\frac{2(h-vt)}{t^2}[/tex] = g
