a projectile is launched straight up from the ground with an initial velocity of 120 ft/s. if acceleration…

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/s$^2$, after about how many seconds will the object reach a height of 200 ft?\n$h(t) = at^2 + vt + h_0$\n1.0 s\n1.7 s\n2.5 s\n8.0 s

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/s$^2$, after about how many seconds will the object reach a height of 200 ft?\n$h(t) = at^2 + vt + h_0$\n1.0 s\n1.7 s\n2.5 s\n8.0 s

Answer

Explanation:

Step1: Substitute given values

Substitute $h(t)=200$, $a=-16$, $v=120$, $h_0=0$ into the height formula: $200 = -16t^2 + 120t + 0$

Step2: Rearrange into standard quadratic form

Bring all terms to one side to get $ax^2+bx+c=0$: $-16t^2 + 120t - 200 = 0$ Divide all terms by $-8$ to simplify: $2t^2 - 15t + 25 = 0$

Step3: Solve quadratic equation

Use quadratic formula $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=2$, $b=-15$, $c=25$: First calculate discriminant: $\Delta = (-15)^2 - 4(2)(25) = 225 - 200 = 25$ Then find roots: $t=\frac{15\pm\sqrt{25}}{4}=\frac{15\pm5}{4}$

Step4: Calculate both possible times

Compute the two solutions: $t_1=\frac{15+5}{4}=\frac{20}{4}=5$ $t_2=\frac{15-5}{4}=\frac{10}{4}=2.5$ The object reaches 200 ft on the way up at 2.5 s and on the way down at 5 s. 2.5 s is among the options.

Answer:

2.5 s