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?

Question

William Hartman

Mathematics

12 February, 00:16

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)

Know the Answer?

Pooch · Answer 1

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