Question

In the following exercises, solve by using methods of factoring, the square root principle, or the quadratic formula.

A bullet is fired straight up from the ground at a velocity of 320 ft/sec. Use the formula $h=-16t^2+v_{0}t$to determine when the bullet will reach 800 feet. Round to the nearest tenth.

Short answer

The bullet will reach 800 feet after $2.9seconds$. It will also pass the height on the way down at $17.1seconds.$

Step-by-step solution

Given information.

The given statement is:

A bullet is fired straight up from the ground at a velocity of 320 ft/sec. Use the formula $h=-16t^2+v_{0}t$to determine when the bullet will reach 800 feet. Round to the nearest tenth.

Substitute the values in the given formula. 

Let the time be $t$seconds.

The height in feet of an object shot upwards in the air with velocity $v_0$after $t$seconds is given by $h=-16t^2+v_{0}t$.

Substitute $h=800,v_0=320$in the given formula.

role="math" localid="1653668207375" $$\begin{gathered} 800=-16t^2+320t \\ -16t^2+320t-800=0 \\ 16t^2-320t+800=0 \\ {t}^2-20t+50=0 \end{gathered}$$

Use the quadratic formula. 

The quadratic formula for the equation $a{x}^2+bx+c=0isx=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$

Now we compare the given equation with the standard form we get

$$\begin{gathered} a=1,b=-20andc=50. \\ t=\frac{-\left(-20\right)\pm \sqrt{{\left(-20\right)}^2-4\left(1\right)\left(50\right)}}{2\left(1\right)} \\ t=\frac{20\pm \sqrt{400-200}}{2} \\ t=10\pm 5\sqrt{2} \\ t=2.9,17.1 \end{gathered}$$

So, the time is$2.9secondsand17.1seconds.$