Gay-Lussac's Law states
P1 / T1 = P2 / T2
So the answer is b
<span>if we assume the origin is at the dropping point and the object is merely dropped and not thrown up or down then y0 = 0 and v0 = 0. The equation reduces to </span>
<span>y = 0 + 0t + ½gt² </span>
<span>y = ½gt² </span>
<span>t = √(2y/g) </span>
<span>in the ft - lb - s system </span>
<span>y = -100 ft </span>
<span>g = -32.2 ft / s² </span>
<span>t = √(2y/g) </span>
<span>t = √(2(-100) / (-32.2)) </span>
<span>t = 2.5 s</span>
Explanation:
an object's gravitational potential energy Eg is m×g×h where:
m=mass
g=9.8m/s²
h=height relative to the closest object below it (because it cannot potentially fall through it
so Eg = 15×9.8×5=735J
