Aprojectile is launched from ground level with an initial velocity of v 0 feet per second. neglecting air resistance, its height in feet t seconds after launch is given by s equals negative 16 t squared plus v 0 t. find the time(s) that the projectile will (a) reach a height of 192 ft and (b) return to the ground when v 0equals128 feet per second.

a) At times t = 2 s and t = 6 s, the projectile will be at a height of 192 ft.

b) The projectile will return to the ground at t = 8 s.  

Explanation:

a) The height of the projectile is given by this equation:

s = -16·t² + 128 f/s·t  (see attached figure)

If the height is 192 ft, then:

192 ft = 16·t² + 128 ft/s· t

0 = -16·t² + 128 ft/s·t - 192 ft

Solving the quadratic equation:

t = 2 and t = 6

At times t = 2 s and t = 6 s, the projectile will be at a height of 192 ft.

b) When the projectile return to the ground, s = 0. Then:

0 = -16·t² + 128 ft/s·t

0 =t(-16·t + 128 ft/s)

t = 0 is the initial point, when the projectile is launched.

-16·t + 128 ft/s = 0

t = -128 ft/s / -16 ft/s² = 8 s

The projectile will return to the ground at t = 8 s.