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)=at^{2}+vt + h_{0}
1.0 s
1.7 s
2.5 s
8.0 s

Explanation:

Step1: Identify the values

Given $a=- 16$, $v = 120$, $h_0=0$ (launched from ground), and $h(t)=200$. Substitute into $h(t)=at^{2}+vt + h_0$. We get $200=-16t^{2}+120t+0$.

Step2: Rearrange to quadratic - form

Rearrange the equation to $16t^{2}-120t + 200 = 0$. Divide through by 8 to simplify: $2t^{2}-15t + 25=0$.

Step3: Use quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 2$, $b=-15$, $c = 25$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-15)^{2}-4\times2\times25=225 - 200 = 25$.

Step4: Solve for t

$t=\frac{15\pm\sqrt{25}}{4}=\frac{15\pm5}{4}$. 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:

C. 2.5 s