Question

calculator
chay throws an object straight up into the air with an initial velocity of 60 ft/s from a platform that is 40 ft above the ground.
use the formula $h(t)=-16t^{2}+v_{0}t + h_{0}$, where $v_{0}$ is the initial velocity and $h_{0}$ is the initial height.
how long will it take the object to hit the ground?
round your answer to the nearest tenth of a second
1.8 s
2.1 s
4.3 s
5.4 s

Explanation:

Step1: Identify values of $v_0$ and $h_0$

$v_0 = 60$, $h_0=40$

Step2: Set up the equation when object hits ground

When the object hits the ground, $h(t)=0$. So the equation becomes $- 16t^{2}+60t + 40=0$. Divide through by -4 to simplify: $4t^{2}-15t - 10=0$.

Step3: Use quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 4$, $b=-15$, $c=-10$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-15)^{2}-4\times4\times(-10)=225 + 160=385$.
Then $t=\frac{15\pm\sqrt{385}}{8}$.
We have two solutions for $t$: $t_1=\frac{15+\sqrt{385}}{8}\approx\frac{15 + 19.62}{8}=\frac{34.62}{8}=4.3$ and $t_2=\frac{15-\sqrt{385}}{8}\approx\frac{15 - 19.62}{8}=\frac{- 4.62}{8}<0$. Since time cannot be negative, we discard the negative - valued solution.

Answer:

$4.3\ s$