Build College Readiness with Targeted 11th Grade MAP Growth Prep

We have a ball being thrown into the air, and its height is described by the equation h(t) = -16t² + 48t + 5. We need to find the maximum height the ball reaches. Think of this like watching a ball go up into the air and then come back down, we want to know how high it gets at its highest point.

This is a quadratic function, which graphs as a parabola. Notice that the coefficient of t² is -16, which is negative. This means the parabola opens downward, like an upside-down U or a frown. The highest point on this curve is called the vertex, and that's exactly where the maximum height occurs. When a parabola opens downward, the vertex represents the maximum value; when it opens upward, the vertex represents the minimum value.

Our equation is h(t) = -16t² + 48t + 5, which is in the standard form at² + bt + c. We need to identify each coefficient:

• a = -16 (the coefficient of t², which tells us the parabola opens downward)
• b = 48 (the coefficient of t, related to the initial velocity)
• c = 5 (the constant term, which represents the initial height of the ball)

To find when the ball reaches its maximum height, we use the vertex formula. The t-coordinate (time) of the vertex is:

t = -b/(2a)

This formula tells us exactly when the maximum or minimum occurs. Let's substitute our values carefully:

t = -b/(2a) t = -(48)/(2 × (-16)) t = -48/(-32) t = 48/32 t = 3/2 t = 1.5 seconds

This tells us that the ball reaches its maximum height exactly 1.5 seconds after it's thrown. This makes sense as the ball goes up for a while, reaches a peak, then starts falling back down.

Now that we know when the maximum occurs (at t = 1.5 seconds), we need to find out what the maximum height actually is. We do this by substituting t = 1.5 back into our original height equation:

h(1.5) = -16(1.5)² + 48(1.5) + 5

Let's work through this calculation step by step:

First, calculate (1.5)²: (1.5)² = 1.5 × 1.5 = 2.25

Next, calculate each term:

• First term: -16(2.25) = -36
• Second term: 48(1.5) = 72
• Third term: 5 (this stays the same)

Finally, add all the terms together: h(1.5) = -36 + 72 + 5 h(1.5) = 36 + 5 h(1.5) = 41 feet

Therefore, the maximum height the ball reaches is 41 feet.

Therefore, answer (B) is correct.

We found this by using the vertex formula to determine that the maximum occurs at t = 1.5 seconds, then substituting that time back into the height equation to find h(1.5) = 41 feet.

Answer (A) 29 ft is incorrect because this value is too low to be the maximum height. Even at the starting moment (t = 0), the ball is already at 5 feet, and it has an upward velocity that will carry it much higher. This answer might result from a calculation error, such as incorrectly finding the vertex time or making arithmetic mistakes when evaluating the function.

Answer (C) 53 ft is incorrect because although this might seem like a reasonable height, it doesn't match what we get when we properly apply the vertex formula and substitute back into the equation. A student might arrive at 53 feet by making an error when squaring 1.5 (perhaps calculating it as 3 instead of 2.25), or by incorrectly adding or subtracting the terms in the final calculation.

Answer (D) 77 ft is incorrect because this value is far too high and unrealistic for this problem. This error might occur if a student forgot to include the negative sign in front of the -16t² term, which would dramatically change the answer. Another way to get this wrong answer would be to simply add up all the numbers you see (16 + 48 + 5) without understanding that you need to substitute a specific value for t and follow the proper calculation steps.