Aprojectile is launched from ground level with a initial velocity of vo feet per second. neglecting air resistance, its height in feet t seconds after launch is given by s= -16t^2 + vot. find the time(s) that the projectile will (a) reach a height of 192 ft and (b) return to the ground when vo = 112 feet per second.

The height at t seconds after launch is
s(t) = - 16t² + V₀t
where V₀ = initial launch velocity.

Part a:
When s = 192 ft, and V₀ = 112 ft/s, then
-16t² + 112t = 192
16t² - 112t + 192 = 0
t² - 7t + 12 = 0
(t - 3)(t - 4) = 0
t = 3 s, or t = 4 s

The projectile reaches a height of  192 ft at 3 s on the way up, and at 4 s on the way down.

Part b:
When the projectile reaches the ground, s = 0.
Therefore
-16t² + 112t = 0
-16t(t - 7) = 0
t = 0 or t = 7 s

When t=0, the projectile is launched.
When t = 7 s, the projectile returns to the ground.

7 s

200 miles.45 x 3 (charge per day times days)135$205$ - 135$ = 70$0.35x = 70divide both sides by 0.35200