<PRE><B>Question 1047598</B>
The arrow is fired 220 ft/s straight up, reaches a maximum height, then falls straight down. The final velocity when it reaches the ground is then used to find the acceleration needed to go from that velocity to zero, over a distance of 2.00 inches, or 2/12 = 1/6 ft. 
:
since the arrow is fired straight up, the y component is  sin(90) * 220 = 220
:
when the arrow reaches max height its velocity is 0, therefore
:
0 = 220 + (-32t) 
:
note that 32 is acceleration of gravity acting downward on the arrow
:
t = 220 / 32 = 6.975 seconds
:
max height of arrow = 220 * 6.975 = 1512.5 feet
:
at max height the arrow has potential energy that equals its kinetic energy at ground 0
:
mgh = (1/2)mv^2
:
32(1512.5) = (1/2)v^2
:
v^2 = 96800
:
v = 311.1 ft/sec
:
now use the equation
:
v(f)^2 = v(0)^2 + 2as
:
0 = 311.1^2 + 2a(1/6)
:
(1/3)*(a) = -96800
:
a = -290400
:
**********************************************************
magnitude of acceleration to stop arrow is 290400 ft/sec^2
**********************************************************
:
</PRE>