A projectile is launched from ground level with an initial velocity of vo feet per second. Neglecting air resistance, its height in feet t seconds after launch is given byS = - 16t2 + vot. Find the time(s) that the projectile will (a) reach a height of 192 ft and (b) return to the ground when Vo = 128 feet per second.(a) Find the time(s) that the projectile will reach a height of 192 ft when vo = 128 feet per second. Select the correct choice below and, if necessary, fill in the answerbox to complete your choice.ChapteOA. seconds (Use a comma to separate answers as needed.)OB. The projectile does not reach 192 feet.entsContents

Where s represent the height, vo the initial velocity and t the time. for part a we want to find the time that the projectile will reach a height of 192 ft if vo= 128 ft/s. So we can use the equation given like this:

[tex]192=-16t^2+128t[/tex]

And we can rewrite the expression like this:

[tex]16t^2-128t+192=0[/tex]

We can divide both sides of the equation by 16 and we got:

[tex]t^2-8t+12=0[/tex]

And we can use the quadratic formula and we got:

[tex]t=\frac{-(-8)\pm\sqrt{(-8)^2-4(1)(12)}}{2(1)}[/tex]

And the solutions for this case are: t=6s and t= 2. So final answer would be t=2,6 sec

Part b

We want to find the tme at which the will return to the ground when vo= 128 ft/s. So we can set up the following equation:

[tex]-16t^2+128t=0[/tex]

And we can rewrite the expression like this:

[tex]16t^2-128t=0[/tex]

We can take common factor 16 t and we got:

[tex]16t(t-8)=0[/tex]

And if we divide both sides of the equation by 16 t we got:

[tex]t-8=0[/tex]

And then the solution for t would be: t=8 sec