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. seconds (use a comma to separate answers as needed.) b. the projectile does not reach 128 feet.

Explanation:

Step1: Substitute the values into the equation

Given \(s = - 16t^{2}+v_{0}t\), \(v_{0}=96\) and \(s = 128\). So the equation becomes \(128=-16t^{2}+96t\).
Rearrange it to the standard quadratic form \(ax^{2}+bx + c=0\): \(16t^{2}-96t + 128 = 0\). Divide through by \(16\) to simplify: \(t^{2}-6t + 8=0\).

Step2: Factor the quadratic equation

For the quadratic equation \(t^{2}-6t + 8 = 0\), we factor it as \((t - 2)(t - 4)=0\).

Step3: Solve for \(t\)

Set each factor equal to zero:
If \(t - 2=0\), then \(t = 2\).
If \(t - 4=0\), then \(t = 4\).

Answer:

A. \(2,4\) seconds