Answer:
4611.58 ft/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 32.174 ft/s²
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 32.174\times 516+0^2}\\\Rightarrow v=182.218\ ft/s[/tex]
[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-182.218^2}{2\times 3.6}\\\Rightarrow a=-4611.58\ ft/s^2[/tex]
Magnitude of acceleration while stopping is 4611.58 ft/s²