Answer:
93750 ft/s²
Explanation:
t = Time taken
u = Initial velocity = 250 ft/s (It is assumed that it is speed of the arrow just when it enter the ground)
v = Final velocity = 0
s = Displacement = 4 in = [tex]\frac{4}{12}=\frac{1}{3}\ feet[/tex]
a = Acceleration
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-250^2}{2\times \frac{1}{3}}\\\Rightarrow a=-93750\ ft/s^2[/tex]
The magnitude of acceleration is 93750 ft/s²
