Question

sandra throws an object straight up into the air with an initial velocity of 38 ft/s from a platform that is 30 ft above the ground. use the formula $h(t) = -16t^2 + v_0t + h_0$, where $v_0$ is the initial velocity and $h_0$ is the initial height. how long will it take for the object to hit the ground? \bigcirc 1 s \bigcirc 2 s \bigcirc 3 s \bigcirc 4 s

Explanation:

Step1: Substitute known values

Substitute $v_0=38$, $h_0=30$, and $h(t)=0$ (ground height) into the formula:
$0 = -16t^2 + 38t + 30$
Multiply both sides by $-1$:
$16t^2 - 38t - 30 = 0$
Simplify by dividing by 2:
$8t^2 - 19t - 15 = 0$

Step2: Apply quadratic formula

For $at^2+bt+c=0$, $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=8$, $b=-19$, $c=-15$:
$$t=\frac{19\pm\sqrt{(-19)^2-4(8)(-15)}}{2(8)}$$
Calculate discriminant:
$\sqrt{361 + 480}=\sqrt{841}=29$

Step3: Solve for valid t

Calculate two solutions:
$t=\frac{19+29}{16}=\frac{48}{16}=3$
$t=\frac{19-29}{16}=\frac{-10}{16}=-0.625$
Discard negative time (invalid).

Answer:

C. 3 s