Ask Question
26 December, 10:20

While on the golf course last weekend Marc hit into the rough landing the ball behind a tall tree. His best option was to get it high enough to get over the tree and hopefully come down in the fairway for his next shot. So with a mighty swing he hit the ball into the air and was surprised to see the ball hit near the top of a 300 foot tall tower that he had not noticed. The formula for this shot is: h (x) = - 16x^2+120 where h is the height of the ball and x is the number of seconds the ball is in the air.

Question 1. How high did Marc actually hit the ball?

Later during that same golf outing, Marc decided to show off by trying to hit the green in one shot. So with his macho swing, he said he hit the green as the ball hung in the air for 15 seconds. The formula for this shot is: h (x) = - 16x^2+200x

Question 2. How long did the ball actually hang in the air?

+3
Answers (1)
  1. 26 December, 10:43
    0
    120 ft at 0 seconds

    Question 2: 12.5 seconds in the air

    Step-by-step explanation:

    h (x) = - 16x^2+120

    We want to find the vertex

    The vertex is at

    h = - b/2a = 0/-32 = 0

    The maximum is at 0

    h (0) = 0+120 = 120

    it occurs at 0 seconds and a height of 120 ft

    h (x) = - 16x^2+200x

    We want to find the zeros of the function

    0 = - 16x^2+200x

    Factor out - 8x

    0 = - 8x (2x - 25)

    Using the zero product property

    -8x=0 2x-25 = 0

    x = 0 2x = 25

    x=25/2

    The ball starts on the ground at 0 seconds and lands at 12.5 seconds
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “While on the golf course last weekend Marc hit into the rough landing the ball behind a tall tree. His best option was to get it high ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers