A projectile 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.

Answer: 7 s