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. seconds (use a comma to separate answers as needed.)
○ b. the projectile does not reach 192 feet.

Explanation:

Step1: Substitute values into the equation

Given \(s = - 16t^{2}+v_{0}t\), \(v_{0}=128\) and \(s = 192\). Substitute these values: \(192=-16t^{2}+128t\).

Step2: Rearrange the equation

Rearrange to get a quadratic equation: \(16t^{2}-128t + 192=0\). Divide throughout by \(16\): \(t^{2}-8t + 12=0\).

Step3: Factor the quadratic equation

Factor \(t^{2}-8t + 12\): \((t - 2)(t - 6)=0\).

Step4: Solve for \(t\)

Set each factor equal to zero: \(t-2 = 0\) gives \(t = 2\); \(t - 6=0\) gives \(t=6\).

Answer:

A. \(2,6\) seconds