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. seconds (use a comma to separate answers as needed.) b. the projectile does not reach 512 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 \(s = 512\), \(v_0=128\), substitute into \(s=-16t^{2}+v_0t\). So, \(512=-16t^{2}+128t\). Rearrange to \(16t^{2}-128t + 512 = 0\). Divide by 16: \(t^{2}-8t + 32=0\).

Step2: Calculate discriminant for part (a)

For quadratic \(at^{2}+bt + c = 0\) (\(a = 1\), \(b=-8\), \(c = 32\)), discriminant \(\Delta=b^{2}-4ac=(-8)^{2}-4\times1\times32=64 - 128=-64<0\).

Step3: Substitute values for part (b)

When the projectile returns to the ground, \(s = 0\), \(v_0 = 128\). Substitute into \(s=-16t^{2}+v_0t\), get \(0=-16t^{2}+128t\). Factor out \(16t\): \(16t(-t + 8)=0\).

Answer:

(a) B. The projectile does not reach 512 feet.
(b) \(8\) seconds.