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_0t ). find the time(s) that the projectile will (a) reach a height of 80 ft and (b) return to the ground when ( v_0 ) is 32 feet per second.
(a) find the time(s) that the projectile will reach a height of 80 ft when ( v_0 = 32 ) feet per second. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. seconds (use a comma to separate answers as needed.)
b. the projectile does not reach 80 feet.
(b) the projectile returns to the ground after second(s). (use a comma to separate answers as needed.)

Explanation:

Step1: Substitute values for part (a)

Given \(s = -16t^{2}+v_{0}t\), \(v_{0}=32\), \(s = 80\).
So, \(80=-16t^{2}+32t\).
Rearrange to \(16t^{2}-32t + 80=0\).
Divide by \(16\): \(t^{2}-2t + 5=0\).

Step2: Calculate discriminant for part (a)

For quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 1\), \(b=-2\), \(c = 5\)), discriminant \(D=b^{2}-4ac\).
\(D=(-2)^{2}-4\times1\times5=4 - 20=-16<0\).
Since discriminant is negative, no real - valued solutions. So, the projectile does not reach 80 feet.

Step3: Substitute values for part (b)

When the projectile returns to the ground, \(s = 0\).
\(0=-16t^{2}+32t\).
Factor out \(-16t\): \(-16t(t - 2)=0\).

Step4: Solve for \(t\) in part (b)

Using zero - product property:
If \(-16t=0\), then \(t = 0\) (initial time).
If \(t - 2=0\), then \(t = 2\).

Answer:

(a) B. The projectile does not reach 80 feet.
(b) \(2\)