Question

A projectile is thrown upward from the ground at 420 ft/sec. Develop the position function using the definition above. At what time is the projectile the farthest from the ground?

Answer 1

Given,
initial velocity (
`v_(i)`) = 420 ft/s

We know,
final velocity (
`v_(f)`) = 0 ft/s

acceleraction (a) = g = -9.8 m/s²
= -32.152231 ft/s²

time (t) = ?

Now,
we know,

`{v_(f)}^2 = {v_(i)}^2 +2ad \\\\ {v_(f)}^2 - {v_(i)}^2=2ad \\\\ d = \frac{{v_(f)}^2 - {v_(i)}^2}{2a} \\\\ d = (0^2 - 420^2)/(2(-32.152231)) \\\\ \boxed{d = 2743.20~ft.}`

Now, lets find time taken to reach the height,


`v_(f) = v_(i) +at \\\\ 0=420 +( -32.152231)(t) \\\\ -420=-32.152231(t) \\\\ (-420)/(-32.152231) = t \\\\ 13.06~s = t \\\\ \boxed{t = 13.06 ~s}`

∴ The porjectile is farthest form the ground at t = 13.06 seconds.
∴ The highest it goes is 2743.20 feet.