Question

a toy rocket is launched straight up into the air with an initial velocity of 60 ft/s from a table 3 ft above the ground. if acceleration due to gravity is -16 ft/s², approximately how many seconds after the launch will the toy rocket reach the ground?
h(t) = at² + vt + h₀
0.05 s
2.03 s
3.80 s
3.70 s

Explanation:

Step1: Identify the values for the formula

Given $a = - 16$, $v = 60$, $h_0=3$, and $h(t)=0$ (when the rocket reaches the ground). So the equation becomes $0=-16t^{2}+60t + 3$.

Step2: Use the quadratic - formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=-16$, $b = 60$, and $c = 3$. First, calculate the discriminant $\Delta=b^{2}-4ac=(60)^{2}-4\times(-16)\times3=3600 + 192=3792$.

Step3: Calculate the values of t

$t=\frac{-60\pm\sqrt{3792}}{-32}=\frac{-60\pm61.58}{-32}$. We have two solutions for $t$: $t_1=\frac{-60 + 61.58}{-32}=\frac{1.58}{-32}\approx - 0.05$ and $t_2=\frac{-60 - 61.58}{-32}=\frac{-121.58}{-32}\approx3.80$. Since time cannot be negative, we take the positive value.

Answer:

C. 3.80 s