Question

An arrow is shot vertically up into the air with an initial vertical velocity of 60 m/s, and its height is given by h=-5t^2+60t where h is in meters and t is in seconds. How high does the arrow go? How long does the arrow stay in flight? Show all work.

Answer 1

In order to find the maximum height, we need to find the vertex of the quadratic equation.

To do so, first let's identify the parameters a, b and c from the standard form:

`\begin{gathered} y=ax^2+bx+c \\ h=-5t^2+60t+0 \\ a=-5,b=60,c=0 \end{gathered}`

Now, we can use the formula below for the vertex x-coordinate:

`t_v_{}=(-b)/(2a)=(-60)/(-10)=6`

Now, calculating the vertex y-coordinate, we have:

`\begin{gathered} h_v=-5t^2_v+60t_v \\ h_v=-5\cdot6^2+60\cdot6 \\ h_v=-180+360 \\ h_v=180\text{ meters} \end{gathered}`

Therefore the maximum height is 180 meters.

Since the vertex occurs for a time of 6 seconds and the time of flight is double the vertex time, the flight time is 12 seconds.