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A diver jumps off a diving board. The diver's height above the water, in meters, is represented by the function h(t)=-49t^2+1.7t+25, where t is the number of seconds since the diver jumped. How high above the water is the diver after 2 seconds? How high above the water is the diver after 3 seconds?

User Elegent
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Final answer:

To find the diver's height above the water, you must plug the given time values into the height function. The diver is 8.8 meters above the water after 2 seconds, and there seems to be an error with the information provided for the height after 3 seconds.

Step-by-step explanation:

To determine how high above the water the diver is after 2 seconds and after 3 seconds, we plug these values into the diver's height function: h(t) = -49t^2 + 1.7t + 25.

  • For t = 2 seconds: h(2) = -49(2)^2 + 1.7(2) + 25 = -49(4) + 3.4 + 25 = -196 + 3.4 + 25 = -167.6 + 25 = -142.6 meters. However, since a negative height does not make sense in this context, it seems there is a miscalculation or typo. In physics, the standard acceleration due to gravity is -9.8 m/s², not -49 m/s². We will correct this and use -4.9 m/s² for the calculation.
  • Corrected for t = 2 seconds: h(2) = -4.9(2)^2 + 1.7(2) + 25 = -4.9(4) + 3.4 + 25 = -19.6 + 3.4 + 25 = 8.8 meters.
  • For t = 3 seconds, using the corrected acceleration: h(3) = -4.9(3)^2 + 1.7(3) + 25 = -4.9(9) + 5.1 + 25 = -44.1 + 5.1 + 25 = -14.0 meters. This result indicates the diver has passed the water surface level and is potentially a typo in the problem statement or suggests the diver has already hit the water before 3 seconds.

The diver's height above the water after 2 seconds is 8.8 meters, and there may be an error with the information provided or a misconception on the diver's height after 3 seconds based on standard physical constants and typical dive scenarios.

User Anarion
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