a toy rocket is launched straight up into the air with an initial velocity of 60 ft/s from a table 3 ft…

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?\nh(t)=at² + vt+h₀\n0.05 s\n2.03 s\n3.80 s\n3.70 s

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?\nh(t)=at² + vt+h₀\n0.05 s\n2.03 s\n3.80 s\n3.70 s

Answer

Explanation:

Step1: Identify the values

We know that $a=- 16$, $v = 60$, $h_0=3$, and we want to find $t$ when $h(t)=0$. 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, $a=-16$, $b = 60$, $c = 3$. First, calculate the discriminant $\Delta=b^{2}-4ac=(60)^{2}-4\times(-16)\times3=3600 + 192=3792$.

Step3: Calculate $t$ values

$t=\frac{-60\pm\sqrt{3792}}{2\times(-16)}=\frac{-60\pm61.58}{-32}$. 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 discard $t_1$.

Answer:

C. 3.80 s