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Ricardo throws a stone off a bridge into a river below. The stone's height (in meters above the water), 2 seconds after Ricardo threw it, is modeled by w(2) = -5( 18)(+1) How many seconds after being thrown will the stone reach its maximum height?

A: 2 seconds
B: 3 seconds
C: 4 seconds
D: 5 seconds

1 Answer

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Final answer:

Without the correct quadratic function, we cannot determine the exact time the stone will reach its maximum height. The stone's maximum height is achieved when its vertical velocity is zero, at the vertex of its parabolic trajectory. We need the correctly notated function to compute the time to reach the maximum height accurately.

Step-by-step explanation:

The stone's maximum height is achieved when its vertical velocity is zero. This occurs at the vertex of the parabolic trajectory of the stone, which in this case is modeled by a quadratic function. Unfortunately, the provided function w(2) = -5( 18)(+1) seems to have typographical errors and does not conform to the standard format of a quadratic equation, so we cannot directly use it to find the time to reach maximum height.However, a correct quadratic function for the height of a projectile thrown upward with initial velocity v from a height h is generally given in the form h(t) = -1/2 g t^2 + v t + h, where g is the acceleration due to gravity (approximately 9.81 m/s^2), t is the time in seconds, and h is the initial height from which the projectile is thrown.

To find the time at which the stone reaches its maximum height, we need to complete the square or use the formula -b/(2a) where a is the coefficient of t^2 and b is the coefficient of t in the quadratic equation.n this question, we have the equation w(2) = -5(18)(+1) which represents the height of the stone above the water, 2 seconds after it was thrown by Ricardo. To find the maximum height, we need to find the time when the derivative of the height function equals zero. So, we need to find the derivative of the function w(t) and set it equal to zero. Then, solve for t.The derivative of w(t) = -5(18)(t+1) is dw/dt = -5(18). Setting this equal to zero, we get -5(18) = 0. Solving for t, we find that t = -1. The stone reaches its maximum height 1 second before being thrown. Therefore, the stone will never reach its maximum height.For this specific scenario, we do not have enough information to determine the exact time the stone will reach its maximum height without the correct function. To answer the student's original question accurately, we would need the correctly notated quadratic equation that models the stone's height as a function of time after being thrown.

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