A thrown ball

A ball thrown up at 80 feet per second from a 40-foot roof reaches 140 feet after 2.5 seconds and lands after about 5.46 seconds.

When is the ball highest, and when does it land?

Two questions first, with a pencil if you have one. A ball thrown straight up from a 40-foot roof at 80 feet per second has height H(t) = −16t² + 80t + 40 after t seconds. When is it highest? When does it hit the ground? Both answers use tools from earlier units. One is a vertex. The other is a root from the quadratic formula. Write your answers down and compare them with the working below.

What the −16, the 80 and the 40 mean

OpenStax uses this example: a ball is thrown upward from the top of a 40-foot-high building at a speed of 80 feet per second, and its height above the ground after t seconds is modeled by H(t) = −16t² + 80t + 40. Each term has a meaning. The constant 40 is the starting height, because at t = 0 only 40 is left. The term 80t is how high the ball would go if it kept its starting speed. The term −16t² is gravity pulling it back, and it grows faster than 80t, which is why the ball eventually turns around and falls.

Where does the 16 come from? Gravity near the ground speeds a falling object up by about 32 feet per second every second; the standard value is 9.80665 m/s², about 32.174 ft/s². Falling distance is half of that times t², and half of 32 is 16. In meters the same model reads h(t) = −4.9t² + vt + h₀. The model ignores air resistance, which is reasonable for a heavy ball over a few seconds but not for a feather.

The top: 140 feet at 2.5 seconds

H(t) is a quadratic with a = −16, so its graph opens downward and the vertex is the highest point. The time of the vertex is t = −b/(2a) = −80/(2 × (−16)) = 80/32 = 2.5 seconds. The height there is H(2.5) = −16(6.25) + 80(2.5) + 40 = −100 + 200 + 40 = 140 feet. So the ball rises for two and a half seconds, peaks at 140 feet and then comes down. Notice that the vertex formula answers a physical question here: when does the ball stop rising?

Landing after about 5.46 seconds

The ball hits the ground when H(t) = 0, that is, −16t² + 80t + 40 = 0. This does not factor nicely, so use the formula: t = (−80 ± √(80² − 4(−16)(40))) / (2(−16)) = (−80 ± √8960) / (−32). The two answers are about 5.458 and about −0.458. A negative time would mean before the throw, which is outside the situation the model describes, so the ball lands after about 5.458 seconds. The discriminant, 8960, is positive, which matches the fact that the graph crosses the time axis.

The same vertex step for a fence and a price

The same vertex idea solves problems with no ball in sight. In another OpenStax example, a gardener has 80 feet of fencing for three sides of a rectangle, with the backyard fence as the fourth side. If each side perpendicular to the existing fence has length L, the long side is 80 − 2L and the area is A(L) = L(80 − 2L) = 80L − 2L². The vertex is at L = −80/(2 × (−2)) = 20, so the best garden is 20 feet by 40 feet, with an area of 800 square feet. A newspaper revenue model in the same section peaks at a price of $31.80 per subscription.