A force of G pounds is required to pull an object weighing W pounds up a ra
1 answer:
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Answer:
Step-by-step explanation:
In order to find the max height the ball reached, we have to complete the square on that quadratic. That will also, conveniently so, give us the number of seconds it will take the ball to reach that max height, that answer to part b. Let's begin to complete the square. Normally, you would move the constant over to the other side of the equals sign, but there is no constant here. The next step is to get the leading coefficient to be a 1, and ours right now is a -15. So we have to factor it out. Here's where we start the process of completing the square.
Next step is to take half the linear term, square it, and add it to both sides. Our linear term is 20/3. Half of 20/3 is 20/6, and 20/6 squared is 400/36.
Because this is an equation, what we add to the left side also has to be added to the right. BUT we didn't just add in 400/36, because we have that -15 out front as a multiplier that refuses to be ignored. What we actually added in was -15(400/36):
The reason we do this is to create a perfect square binomial on the left which will serve as the number of seconds, h, in the vertex (h, k), where h is the number of seconds it takes the ball to reach its max height, k. Simplifying both sides then gives us:
Finally, we will move the right side over by the left and set the quadratic back equal to h(t):
and from that you can determine that the vertex is
.
The answer to a. is vound in the second number of our vertex: k, the max height. The max of the golf ball was 500/3 feet or 166 2/3 feet.
Part b is found in the first number of the vertex: h, the number of seconds it took the golf ball to reach that max height. The time it took was 3 1/3 seconds.
Part c. is to state the domain (the time) and the range (the height) of the ball.
Domain is
D: {x | 0 ≤ x ≤ 3 1/3} and
Range is
R: {y | 0 ≤ y ≤ 166 2/3}