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 512 ft and (b) return to the ground when ( v_0 ) is 128 feet per second. (a) find the time(s) that the projectile will reach a height of 512 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. ( square ) seconds (use a comma to separate answers as needed.) b. the projectile does not reach 512 feet.

Explanation:

Step1: Substitute values into the equation

Given \(s = - 16t^{2}+v_{0}t\), \(v_{0}=128\) and \(s = 512\). Substitute these values into the equation:
\(512=-16t^{2}+128t\)

Step2: Rearrange the equation

Rearrange the equation to the standard quadratic form \(ax^{2}+bx + c=0\).
\(16t^{2}-128t + 512=0\). Divide through by \(16\) to simplify: \(t^{2}-8t + 32=0\).

Step3: Use the quadratic formula

The quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). For the equation \(t^{2}-8t + 32 = 0\), where \(a = 1\), \(b=-8\) and \(c = 32\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-8)^{2}-4\times1\times32=64 - 128=-64\)

Answer:

Since \(\Delta=-64<0\), the equation \(t^{2}-8t + 32 = 0\) has no real solutions. So the answer is B. The projectile does not reach 512 feet.