Terry throws a ball straight up. The initial height of the ball is 3 feet and has an initial velocity of 32 feet per second. The function is modeled by: h(t) = -16t2 + 32t + 3 a) What are the coordinates of the vertex of the parabola? b) What do the coordinates of the vertex of the parabola mean in the context of this problem? -How high off the ground will the ball be after 2 seconds? Show your work -What is the height of the ball at 0.5 seconds and 1.5 seconds after being thrown? Why is the height the same at .5 seconds and 1.5 seconds?
a) How long is the ball in flight? Round answer to nearest hundredth. b) Which formula, process, etc. did you use to solve this problem?
The vertex of a parabola can be found by solving for -b/2a. So in this case, -b = -32 and 2a = -32, meaning -b/2a = 1. This is the x-coordinate for your vertex. To get the coordinates, you plug this value in for x and solve for y. so -16(1)^2 +32(1) + 3 = 19. So the coordinates would be (1,19). These coordinates represent the maximum height the ball will reach before falling back to the ground. The next part, how high will the ball be after 2 seconds, is solved by plugging in 2 for t in the equations. -16(2)^2 + 32(2) + 3 = -64 + 64 + 3 = 3. So the height will be 3 feet (where you started from basically). As for why the height is the same at .5 and 1.5, its because parabolas are symmetrical. The line that travels vertically through the vertex of a parabola is called the axis of symmetry. This line essentially separates the parabola into 2 identical sides. So .5 and 1.5 are equal distances away from the vertex at 1, theyre going to have the same values. In order to find out how long the ball is in flight, you need to factor the qudratic equation and set it equal to 0. In this case, you would need the quadratic formula it looks like. Do you know the quadratic formula and do you think you could solve for t on your own? If not I can help, but those are the first parts to the problem. Sorry for the long explanation.
Thank you so much!
can you help on the last part
Well, the quadratic formula is: *drawing*
|dw:1374861500377:dw| in your problem, a = -16 b = 32 and c = 3. From there you are just plugging in numbers and solving, doing your best to be careful of course.
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