Question

Supposed Aven drops a ball off the top of a 10 foot pool slide, and the ball follows the projectile h(t) = -16t^2+6 , where t is the time in seconds, and h is the height of the ball. Her friend Riley needs to catch the ball between 2 feet and 5 feet off the top of the water. Between what two times should Riley try to catch the ball?

Answer 1

The answer is very simple. .

`h(t)=-16t^2+6\text{ }`

In this equation the heights values are substituted to determine the times.

For h = 2 feet

`\begin{gathered} 2=-16t^2\text{ + 6 } \\ -16t^2\text{ = 2 - 6} \\ -16t^2\text{ = -4} \end{gathered}`
`\begin{gathered} t^2\text{ = }\frac{-4}{-16\text{ }} \\ t^2\text{ = }(1)/(4) \end{gathered}`
`t\text{ = }\sqrt[]{(1)/(4)}\text{ = }0.5\text{ seconds}`

That would be the first time corresponding to the height of 2 feet.

We still need to calculate the time for the other height.

For h = 5 feet

`-16t^2+6\text{ =5}`
`\begin{gathered} -16t^2\text{ = 5 -6 } \\ -16t^2\text{ = -1} \\ t^2\text{ = }(1)/(16) \\ t\text{ = }\sqrt[]{(1)/(16)}\text{ = }0.25\text{ seconds} \end{gathered}`

The answer is: Riley should try to catch the ball between 0.25 and 0.5 seconds.