Answer:
3 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 32 ft/s²
[tex]v=u+at\\\Rightarrow 0=u-9.8\times t\\\Rightarrow \frac{-112}{-32}=t\\\Rightarrow t=3.5 \s[/tex]
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=112\times 3.5+\frac{1}{2}\times -32\times 3.5^2\\\Rightarrow s=196\ ft[/tex]
174 feet from the ground is 174-14 = 160 ft from the launch area
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -32\times 160+112^2}\\\Rightarrow v=48\ ft/s[/tex]
[tex]v=u+at\\\Rightarrow 0=48-32t\\\Rightarrow t=\frac{-48}{-32}=1.5\ s[/tex]
When the arrow will reach the 160 ft point while returning the initial velocity becomes equal to the final velocity which means the time taken to come down also becomes equal.
Hence, the arrow will be 1.5+1.5 = 3 seconds above a height of 174 ft from the ground.
