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For a physics experiment, the class drops a golf ball off a bridge toward the pavement below. The bridge is 75 feet high. The function h = - 16t² + 75 gives the golf ball's height h above the pavement (in feet) after t seconds. Use the graph of the function on the right. After seconds does the golfball hit the pavement

2 Answers

2 votes

Answer:

2.17 seconds

Explanation:

User BlamKiwi
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4 votes

Answer:

To find out when the golf ball hits the pavement (when the height is 0 feet), we can set h = 0 in the equation h = -16t^2 + 75 and solve for t:

0 = -16t^2 + 75

16t^2 = 75

t^2 = 75/16

t = sqrt(75/16)

The square root of (75/16) is approximately 1.861 seconds, so the golf ball hits the pavement after approximately 1.861 seconds.

User Concat
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