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 192 ft and (b) return to the ground when $v_0 = 128$ feet per second. (a) find the time(s) that the projectile will reach a height of 192 ft when $v_0 = 128$ feet per second. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. 2,6 seconds (use a comma to separate answers as needed.) b. the projectile does not reach 192 feet. (b) the projectile returns to the ground after 2 second(s).

Explanation:

Step1: Substitute \(v_0 = 128\) into the height formula

Given \(s=-16t^{2}+v_{0}t\), when \(v_0 = 128\), the formula becomes \(s=-16t^{2}+128t\).

Step2: Solve for \(t\) when \(s = 192\) (part a)

Set \(-16t^{2}+128t=192\).
First, divide the entire equation by \(-16\): \(t^{2}-8t + 12=0\).
Factor the quadratic equation: \((t - 2)(t - 6)=0\).
Using the zero - product property \(t-2=0\) or \(t - 6=0\), so \(t = 2\) or \(t=6\).

Step3: Solve for \(t\) when \(s = 0\) (part b)

Set \(-16t^{2}+128t=0\).
Factor out \(-16t\): \(-16t(t - 8)=0\).
Using the zero - product property \(-16t=0\) or \(t - 8=0\). Since \(t = 0\) is the launch time, the time when it returns to the ground is \(t = 8\).

Answer:

(a) \(2,6\) seconds; (b) \(8\) seconds.