185k views
1 vote
Can you help me please?A. How can Marc provide proof that his mighty shot actually hung in the air for 15 seconds? Or is this just another one of his lies?B. How long did the ball actually hang in the air?

Can you help me please?A. How can Marc provide proof that his mighty shot actually-example-1

1 Answer

6 votes

The given formula for Marc's shot is:


h(x)=-16x^2+200x

a. To prove that the shot actually hung in the air for 15 seconds, we need to replace x=15 in the formula and solve for h, as follows:


\begin{gathered} h(15)=-16(15)^2+200(15) \\ h(15)=-16*225+3000 \\ h(15)=-3600+3000 \\ h(15)=-600 \end{gathered}

As the height is negative, it means after 15 seconds the ball already hit the ground, because the ground is located at h=0. Then this result proves that this is just another one of Marc's lies.

b. To find how long the ball actually hung in the air, we need to find the x-values that makes h=0, as follows:


0=-16x^2+200x

We have a polynomial in the form: ax^2+bx+c=0, where a=-16, b=200 and c=0.

We can use the quadratic formula to solve for x:


\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-200\pm\sqrt[]{(200)^2-4(-16)(0)}}{2(-16)} \\ x=\frac{-200\pm\sqrt[]{40000+0}}{-32} \\ x=\frac{-200\pm\sqrt[]{40000}}{-32} \\ x=(-200\pm200)/(-32) \\ x=(-200+200)/(-32)=(0)/(-32)=0\text{ and }x=(-200-200)/(-32)=(-400)/(-32)=12.5 \end{gathered}

Then the two x-values are x=0 and x=12.5.

The starting time is 0 and the end time when the ball hit the ground is x=12.5.

The ball actually hung in the air 12.5 seconds.

User Zimmryan
by
3.9k points