Question

The function y = -0.018 + 0.56t models the height y in feet of your pet frog's jump t seconds after it jumps. How far did the frog jump? How high did it go?AThe frog jumped about 15.56 ft. far and about 4.36 ft. high.The frog jumped about 4.36 ft. far and about 31.11 ft. high.BхThe frog jumped about 31.11 ft. far and about 4.36 ft. high.СхThe frog jumped about 31.11 ft. far and about 15.56 ft. high.Dх

Answer 1

The Solution.

Given the function below:

$y=-0.018t^2+0.56t$

We differentiate the function with respect to t and equate it to zero.

$\frac{df}{dx}=2(-0.018)t^{2-1}+1(0.56)t^{1-1}=0$$\begin{gathered} -0.036t+0.56=0 \\ \text{Solving for t, we get} \\ -0.036t=-0.56 \\ \text{Dividing both sides by -0.036, we get} \\ t=\frac{-0.56}{-0.036}=15.555\approx15.56\text{ fe}ets \end{gathered}$

So, the frog jumped about 15.56 feet far.

To find how high the frog jumped.

We shall substitute 15.56 for t in the given function.

$undefined$