Answer:
Explanation:
Use the equation
[tex]v^2=v_0^2+2a[/tex]Δx
In this dimension (the y dimension...the only one we have to care about for this problem), here's what we know:
a = -32 ft/s/s
v₀ = 0 m/s (since someone was holding the stone still before it was dropped)
v = 136 ft/s
Δx = ??
Filling in:
[tex]136^2=0^2+2(-32)[/tex]Δx so
Δx = [tex]\frac{136^2}{2(-32)}[/tex] so
Δx = -289 feet (negative because the stone dropped that many feet below the point from which it was dropped, but you would answer a height question with the positive of this number. If you were asked how far the stone dropped, it would be negative; if you're asked how tall the cliff is, that would be positive)
