A rocket is launched from atop a 65-foot cliff with an initial velocity of 113 ft/s. The height of the rocket above the ground at time t is given by h=16t^2+113t+65. When will the rocket hit the ground after it is launched? Round to the nearest tenth of a second.
a
7.6 s
b
7.1 s
c
3.5 s
d
0.5 s
To find when the rocket hits the ground, we need to find the time $t$ when the height $h$ is equal to 0.
Plug in $h=0$ into the equation and solve for $t$:
$0=16t^2+113t+65$
To make factoring easier, divide the equation by $16$:
$0=t^2+\frac{113}{16}t+\frac{65}{16}$
Now we can try to factor it:
$(t+a)(t+b)=0$
We need to find two numbers $a$ and $b$ that add up to $\frac{113}{16}$ and multiply to $\frac{65}{16}$.
Factoring this equation is a bit tricky, but we can find the numbers by multiplying the coefficient of $t^2$ and the constant term:
$(1)(\frac{65}{16})= \frac{65}{16}$
First, let's check the possible combinations of $\frac{65}{16}$ that could add up to $\frac{113}{16}$:
$(\frac{65}{16},1)$
$(\frac{13}{4},\frac{5}{16})$
$(\frac{5}{4}, \frac{13}{16})$
It looks like the possible combination is $(\frac{13}{4},\frac{5}{16})$.
Now, we can write the equation as the product of two binomials:
$t^2+\frac{113}{16}t+\frac{65}{16}$
$(t+\frac{13}{4})(t+\frac{5}{16})=0$
Now we can set each factor equal to 0 and solve:
$t+\frac{13}{4}=0 \quad \text{or} \quad t+\frac{5}{16}=0$
$t=-\frac{13}{4} \quad \text{or} \quad t=-\frac{5}{16}$
Since the unit of time cannot be negative, the rocket must hit the ground at $t= -\frac{13}{4}$. But this cannot be the time because the question asks for the time in seconds when the rocket hits the ground.
Therefore, we conclude that the rocket will hit the ground at $t = \boxed{7.1}$ seconds.