Question

a projectile is launched straight up from the ground with an initial velocity of 120 ft/s. if acceleration due to gravity is - 16 ft/s², after about how many seconds will the object reach a height of 200 ft? h(t)=at² + vt + h₀ 1.0 s 1.7 s 2.5 s 8.0 s

Explanation:

Step1: Substitute values into the formula

Given \(a=- 16\), \(v = 120\), \(h_0=0\) (launched from the ground), and \(h(t)=200\). Substitute into \(h(t)=at^{2}+vt + h_0\), we get \(200=-16t^{2}+120t+0\), which simplifies to \(16t^{2}-120t + 200 = 0\). Divide through by \(8\) to get \(2t^{2}-15t + 25=0\).

Step2: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 2\), \(b=-15\), \(c = 25\)), use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-15)^{2}-4\times2\times25=225 - 200=25\).
Then \(t=\frac{15\pm\sqrt{25}}{4}=\frac{15\pm5}{4}\).
We have two solutions: \(t_1=\frac{15 + 5}{4}=\frac{20}{4}=5\) and \(t_2=\frac{15-5}{4}=\frac{10}{4}=2.5\).

Answer:

\(2.5\ s\) (corresponding to the option \(2.5\ s\))