Calculating Ball Height and Time with Quadratic Formula

[mathdad]

A ball is thrown straight upward. Suppose that the height of the ball at time t is h = -16t^2 + 96t, where h is in feet and t is in seconds, with t = 0 corresponding to the instant the ball is first tossed.

A. How long does it take for the ball to land?

To do A, I must let h = 0 and solve for t, right?

B. At what time is the height 80 feet? Why does B have two answers?

To do B, I must let h = 80 and solve for t, right?

 

[Prove It]

A ball is thrown straight upward. Suppose that the height of the ball at time t is h = -16t^2 + 96t, where h is in feet and t is in seconds, with t = 0 corresponding to the instant the ball is first tossed.

A. How long does it take for the ball to land?

To do A, I must let h = 0 and solve for t, right?

B. At what time is the height 80 feet? Why does B have two answers?

To do B, I must let h = 80 and solve for t, right?

Yes that is what you need to do for both.

 

[mathdad]

Good. I will answer both parts tonight.

 

[MarkFL]

Since you are asked multiple questions regarding finding the time when the ball is a certain height, what I recommend is solving the given relation between height $h$ and time $t$ for $t$, so that you then have a formula to use. We are given:

\(\displaystyle h=-16t^2+96t\)

Arrange this as:

\(\displaystyle 16t^2-96t+h=0\)

Now use the quadratic formula to obtain:

\(\displaystyle t=\frac{12\pm\sqrt{144-h}}{4}\)

Now it's just a matter of plugging in any given height to find the time when that height occurs, rather than having to solve a quadratic equation every time a new height is introduced. We can also easily see that the maximum height is 144 and occurs at time $t=3$. :D

 

[mathdad]

Since you are asked multiple questions regarding finding the time when the ball is a certain height, what I recommend is solving the given relation between height $h$ and time $t$ for $t$, so that you then have a formula to use. We are given:

\(\displaystyle h=-16t^2+96t\)

Arrange this as:

\(\displaystyle 16t^2-96t+h=0\)

Now use the quadratic formula to obtain:

\(\displaystyle t=\frac{12\pm\sqrt{144-h}}{4}\)

Now it's just a matter of plugging in any given height to find the time when that height occurs, rather than having to solve a quadratic equation every time a new height is introduced. We can also easily see that the maximum height is 144 and occurs at time $t=3$. :D

Nicely done. You created an equation similar to the quadratic formula.