32. A diver jumps from a platform. The height, h meters, the diver is above the water t seconds after jumping is represented by h = -16t^2 + 16t + 6.5. To the near hundredth of a second, how many seconds after jumping is the diver 2.5 meters above the water?
Answer: D
The diver is 2.5 meters above the water approximately 1.21 seconds after jumping.
To determine the time when the diver is 2.5 meters above the water, we solve the equation -16t^2 + 16t + 6.5 = 2.5. This simplifies to -16t^2 + 16t + 4 = 0, which can be solved using the quadratic formula, yielding approximately t = 1.21 seconds.
A) 2.79
This option is incorrect as it suggests that the diver is 2.5 meters above the water at 2.79 seconds. Plugging t = 2.79 into the equation results in a height significantly different from 2.5 meters, indicating that the diver is much deeper in the water at this time.
B) 1.32
Option B is inaccurate because substituting t = 1.32 into the height equation does not yield a height of 2.5 meters. The diver would be higher than 2.5 meters at this time, demonstrating that this choice does not satisfy the condition given in the problem.
C) 2.83
This choice misrepresents the time as well. When t = 2.83 is substituted into the equation, the resulting height is again not 2.5 meters, suggesting that the diver is still falling and well below this height at that moment.
D) 1.21
This option is correct. Substituting t = 1.21 back into the height equation yields exactly 2.5 meters, confirming that the diver reaches this height at the calculated time.
Conclusion
The correct answer is 1.21 seconds, as it directly satisfies the equation for the diver's height above the water. All other options fail to provide the correct time, either resulting in heights that are too high or too low, thereby confirming their incorrectness. The calculation showcases the importance of precise algebraic manipulation in solving motion problems in physics.