Question

An arrow is shot vertically upward at a rate of 220 feet per second. Use the projectile formula $h=-16t^2+v_{o}t$, to determine when height of the arrow will be 400 feet.

Short answer

The arrow goes up and then fall down. As the arrow goes up it will reach 400 feet after approximately$2.2seconds$. It will also pass the height on the way down at$11.6seconds$

Step-by-step solution

Step 1. Given Information

The given data is the formula of projectile$h=-16t^2+v_{o}t$

Step 2. Calculation

Substitute $v_o=220,h=400$in the given projectile formula and solve the equation for time.

$$\begin{gathered} 400=-16t^2+220t \\ -16t^2+220t-400=0 \\ 16t^2-220t+400=0 \end{gathered}$$

On comparing with the standard form, we have, $a=16,b=-220,c=400$

Now, use the quadratic formula to find the value of time.

$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ t=\frac{-(-220)\pm \sqrt{{(-220)}^2-4\left(16\right)\left(220\right)}}{2\left(16\right)} \\ t=\frac{220\pm \sqrt{22800}}{32} \\ t\approx 11.6,2.2 \end{gathered}$$

Thus, the arrow goes up and reach 400 feet after approximately $2.2s$and pass the height on the way down at$11.6s$