Aprojectile is launched from ground level with an initial velocity of v0 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 ts=−16t2+v0t. find the time(s) that the projectile will (a) reach a height of 192 ft and (b) return to the ground when v 0v0equals=112 feet per second.

Step-by-step explanation:

It is given that, a projectile is launched from ground level. Its height as a function of time is given by :

...............(1)

Where

v₀ is the initial velocity of the projectile,

(a) Let t is the time taken by the projectile to reach a height of 192 ft. Equation (1) becomes :

Aftyer solving the above equation we get :

t = 3 seconds (on the way up)

and t = 4 seconds (on the way down)

(b) When the projectile returns to the ground, h(t) = 0                  

On solving the above quadratic equation, t = 7 seconds

Hence, this is the required solution.    

the answer is 62.8 cubic units