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:

1. First, identify the values in the height - time formula \(h(t)=at^{2}+vt + h_{0}\):

• The acceleration \(a=- 16\) ft/s² (negative because it acts in the opposite direction of motion), the initial velocity \(v = 120\) ft/s, and the initial height \(h_{0}=0\) (launched from the ground), and we want to find \(t\) when \(h(t)=200\) ft.
• Substitute these values into the formula: \(200=-16t^{2}+120t + 0\).
• Rearrange the equation to get a quadratic equation in standard form: \(16t^{2}-120t + 200 = 0\). Divide through by 8 to simplify: \(2t^{2}-15t + 25=0\).

1. Then, use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for a quadratic equation \(ax^{2}+bx + c = 0\).

• For the equation \(2t^{2}-15t + 25 = 0\), \(a = 2\), \(b=-15\), and \(c = 25\).
• 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 \(t_2=\frac{15 - 5}{4}=\frac{10}{4}=2.5\).
• The smaller - valued \(t = 2.5\) s represents the time when the projectile is going up and reaches a height of 200 ft, and the larger - valued \(t = 5\) s represents the time when the projectile is coming back down and reaches a height of 200 ft.

Answer:

C. 2.5 s