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/s², after about how many seconds will the object reach a height of 200 ft?\nh(t)=at² + vt+h₀\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/s², after about how many seconds will the object reach a height of 200 ft?\nh(t)=at² + vt+h₀\n1.0 s\n1.7 s\n2.5 s\n8.0 s

Answer

Explanation:

Step1: Identify the values

We know that $a=- 16$ (acceleration due to gravity), $v = 120$ (initial velocity), $h_0=0$ (launched from ground), and $h(t)=200$. Substitute into $h(t)=at^{2}+vt + h_0$. So we get $200=-16t^{2}+120t+0$.

Step2: Rearrange the equation

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

Step3: 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 $t=\frac{15\pm\sqrt{25}}{4}=\frac{15\pm5}{4}$.

Step4: Find the two solutions for t

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