Given,
initial velocity ([tex] v_{i} [/tex]) = 420 ft/s
We know,
final velocity ([tex] v_{f} [/tex]) = 0 ft/s
acceleraction (a) = g = -9.8 m/s²
= -32.152231 ft/s²
time (t) = ?
Now,
we know,
[tex] {v_{f}}^2 = {v_{i}}^2 +2ad \\\\ {v_{f}}^2 - {v_{i}}^2=2ad \\\\ d = \frac{{v_{f}}^2 - {v_{i}}^2}{2a} \\\\ d = \frac{0^2 - 420^2}{2(-32.152231)} \\\\ \boxed{d = 2743.20~ft.}[/tex]
Now, lets find time taken to reach the height,
[tex] v_{f} = v_{i} +at \\\\ 0=420 +( -32.152231)(t) \\\\ -420=-32.152231(t) \\\\ \frac{-420}{-32.152231} = t \\\\ 13.06~s = t \\\\ \boxed{t = 13.06 ~s}[/tex]
∴ The porjectile is farthest form the ground at t = 13.06 seconds.
∴ The highest it goes is 2743.20 feet.
