MHB Calculating Ball Height and Time with Quadratic Formula

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To determine how long it takes for a ball thrown upward to land, set the height equation h = -16t^2 + 96t to zero and solve for t. The quadratic formula can be rearranged to t = (12 ± √(144 - h)) / 4, allowing for quick calculations of time at various heights. For the height of 80 feet, there are two solutions because the ball reaches this height on its way up and again on its way down. The maximum height of the ball is 144 feet, occurring at t = 3 seconds. This approach simplifies finding the time for any specified height without repeatedly solving the quadratic equation.
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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?
 
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RTCNTC said:
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.
 
Good. I will answer both parts tonight.
 
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:

$$h=-16t^2+96t$$

Arrange this as:

$$16t^2-96t+h=0$$

Now use the quadratic formula to obtain:

$$t=\frac{12\pm\sqrt{144-h}}{4}$$

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
 
MarkFL said:
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:

$$h=-16t^2+96t$$

Arrange this as:

$$16t^2-96t+h=0$$

Now use the quadratic formula to obtain:

$$t=\frac{12\pm\sqrt{144-h}}{4}$$

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.
 
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