Question

a projectile 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=-16t^{2}+v_{0}t ). find the time(s) that the projectile will (a) reach a height of 128 ft and (b) return to the ground when ( v_{0}=96 ) feet per second
(a) find the time(s) that the projectile will reach a height of 128 ft when ( v_{0}=96 ) feet per second. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. 2,4 seconds (use a comma to separate answers as needed.)
b. the projectile does not reach 128 feet.
(b) the projectile returns to the ground after ( square ) second(s).

Explanation:

Step1: Substitute values into the height formula for part (a)

Given \(h=-16t^{2}+v_{0}t\), \(v_{0} = 96\), and \(h = 128\). Substitute these values:
\(128=-16t^{2}+96t\)
Rearrange to standard quadratic form \(ax^{2}+bx + c=0\):
\(16t^{2}-96t + 128 = 0\)
Divide through by \(16\):
\(t^{2}-6t + 8=0\)

Step2: Factor the quadratic equation

Factor \(t^{2}-6t + 8\):
\((t - 2)(t - 4)=0\)
Set each factor equal to zero:
\(t-2=0\) gives \(t = 2\)
\(t - 4=0\) gives \(t=4\)

Step3: Substitute values into the height formula for part (b)

When the projectile returns to the ground, \(h = 0\). With \(v_{0}=96\), the equation is \(0=-16t^{2}+96t\)
Factor out \(-16t\):
\(0=-16t(t - 6)\)
Set each factor equal to zero:
\(-16t=0\) gives \(t = 0\) (launch time, not the return - to - ground time)
\(t-6=0\) gives \(t = 6\)

Answer:

(a) \(2,4\) seconds
(b) \(6\) seconds