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 1760 ft with an initial velocity of 16 ft/ s. how long will it take for the object to reach the ground?

Explanation:

Step1: Substitute values into formula

Given $h = - 16t^{2}+v_{0}t + h_{0}$, $v_{0}=16$, $h_{0}=1760$, and when the object reaches the ground $h = 0$. So we have the equation $0=-16t^{2}+16t + 1760$.

Step2: Simplify the equation

Divide the entire equation by -16: $t^{2}-t - 110=0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b=-1$, $c=-110$), we use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-1)^{2}-4\times1\times(-110)=1 + 440 = 441$. Then $t=\frac{1\pm\sqrt{441}}{2}=\frac{1\pm21}{2}$.

Step4: Find valid solutions for t

We get two solutions: $t_{1}=\frac{1 + 21}{2}=11$ and $t_{2}=\frac{1-21}{2}=-10$. Since time $t\geq0$, we discard the negative - valued solution.

Answer:

$11$