A water fountain is designed to shoot a stream of water in the shape of a parabolic arc. The equation of the parabola is given by ℎ(𝑡)=−0.5𝑡2+4𝑡+1h(t)=−0.5t 2 +4t+1, where ℎ(𝑡)h(t) represents the height of the water stream in meters and t  represents the time in seconds since the water was shot. Answer the following questions.Determine the maximum height reached (in meters).

ProblemWhen a sprinkler is installed in the ground, the spray of water goes up and falls in the pattern of a parabola. The height, in inches, of a spray of water is given by the equation h(x)=160x−16x2ℎ(𝑥)=160𝑥−16𝑥2 where x𝑥 is the number of feet away from the sprinkler head the spray is. What is the height of the spray 22 feet away from the sprinkler head? Solutionh(2)=ℎ(2)= After 22 seconds, the height of the spray is inches.How many feet along the ground away from the sprinkler head will the spray reach a maximum height? feetWhat is the maximum height of the water spray? inchesHow many feet away from the sprinkler head will the water hit the ground again? feet

ACTIVITY 3: How high and how farA fire fighter aims a fire hose upward, toward a fire in a skyscraper. The water leaving thehose has a velocity of 32.0 m/s. If the fire fighter holds the hose at an angle of 78.5°, whatis the maximum height and range of the water stream? (Write your answer in yournotebook)Note: The water droplets leaving the hose can be treated as projectiles

1. Jason jumped off a cliff into the ocean in Acapulco while vacationing with some friends. His height as a function of time could be modeled by the function h(t) = -16t² + 16t + 480, where t is the time in seconds and h is the height in feet.a. How long did it take for Jason to reach his maximum height?b. What was the highest point that Jason reached?c. Jason hit the water after how many seconds?

A ball is thrown upwards from a bridge into a river below. The height of the ball above the water, in meters, can be modelled by the quadratic equation , where  is the time in seconds since the ball was thrown. Determine at what time(s) is the ball  meters above the water. Interpret the solutions.

Let’s revisit our archery problem from earlier with a little more information.Andrew is an avid archer. He launches an arrow that takes a parabolic path. The equation of the height of the arrow with respect to time is y=−4.9x2+48x𝑦=−4.9𝑥2+48𝑥, where y𝑦 is the height of the arrow in meters above Andrew’s bow and x𝑥 is the time in seconds since Andrew shot the arrow. Find how long it takes the arrow to come back to a height even with his bow height.Let’s graph the equation by making a table of values. Remember it is helpful to start with finding the vertex so that we can know what values to use for x𝑥.y=−4.9x2+48x𝑦=−4.9𝑥2+48𝑥Find the x𝑥-value of the vertex by using the formula −b2a−𝑏2𝑎.−482(−4.9)≈4.9−482(−4.9)≈4.9Note: We used the ≈≈ symbol instead of the == sign because we are having to round this value. Remember we aren’t going to get as many “pretty” numbers with real world problems.)Now we’ll substitute that into our equation to find the y𝑦-value of our vertex.y=−4.9(𝑦=−4.9( )2+48()2+48( )=117.6)=117.6Our vertex is approximately (( ,, )).We’ll create our table of values centered around our vertex (but will round to 55 for that spot in our table).x𝑥y=−4.9x2+48x𝑦=−4.9𝑥2+48𝑥00y=−4.9(0)2+48(0)=0𝑦=−4.9(0)2+48(0)=011y=−4.9(1)2+48(1)=43.1𝑦=−4.9(1)2+48(1)=43.122y=−4.9(2)2+48(2)=76.4𝑦=−4.9(2)2+48(2)=76.433y=−4.9(3)2+48(3)=99.9𝑦=−4.9(3)2+48(3)=99.944y=−4.9(4)2+48(4)=113.6𝑦=−4.9(4)2+48(4)=113.655y=−4.9(5)2+48(5)=117.6𝑦=−4.9(5)2+48(5)=117.666y=−4.9(6)2+48(6)=111.6𝑦=−4.9(6)2+48(6)=111.677y=−4.9(7)2+48(7)=95.9𝑦=−4.9(7)2+48(7)=95.988y=−4.9(8)2+48(8)=70.4𝑦=−4.9(8)2+48(8)=70.499y=−4.9(9)2+48(9)=35.1𝑦=−4.9(9)2+48(9)=35.11010y=−4.9(10)2+48(10)=−10𝑦=−4.9(10)2+48(10)=−10Here’s the graph of the function.The roots of the function are approximately x=0𝑥=0 sec and x=9.8𝑥=9.8 sec. The first root tells us that the height of the arrow was 00 meters above his bow after seconds (right before Andrew first shot it) which makes sense. The second root says that it takes approximately seconds for the arrow to return to the height of the bow.Remember that we already calculated our vertex, so we don’t need to inspect our graph to find it, but we can use it to confirm it. We can interpret our vertex to mean that at approximately seconds, the arrow reached a maximum/minimum height of meters.CheckQuestion 7