Question

for the following question, use the function h = - 16t² + v₀t + h₀, where h is the height of the object after time t, v₀ is the initial velocity, and h₀ is the initial height. an object is thrown upward from a height of 704 ft with an initial velocity of 112 ft/ s. how long will it take for the object to reach the ground? answer 2 points

Explanation:

Step1: Identify the values of variables

Given $h_0 = 704$, $v_0=112$, and when the object reaches the ground $h = 0$. Substitute into $h=-16t^{2}+v_0t + h_0$ to get $0=-16t^{2}+112t + 704$.

Step2: Simplify the equation

Divide the entire equation by -16: $t^{2}-7t - 44=0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b=-7$, $c=-44$), use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-7)^{2}-4\times1\times(-44)=49 + 176 = 225$. Then $t=\frac{7\pm\sqrt{225}}{2}=\frac{7\pm15}{2}$.

Step4: Find the valid solution

We get two solutions for $t$: $t_1=\frac{7 + 15}{2}=\frac{22}{2}=11$ and $t_2=\frac{7-15}{2}=\frac{-8}{2}=-4$. Since time $t\geq0$, we discard the negative - valued solution.

Answer:

$11$