Answer :
Let's solve the problem of finding the maximum height attained by a ball thrown directly upward with a velocity of [tex]\(46 \, \text{ft/s}\)[/tex]. The height [tex]\(y\)[/tex] of the ball as a function of time [tex]\(t\)[/tex] is given by the equation:
[tex]\[ y = 46t - 16t^2 \][/tex]
To find the maximum height, we need to determine the highest point of this parabolic trajectory. We can do this via the following steps:
1. Determine the time at which the maximum height occurs:
The maximum height of a parabola that opens downward, given by a quadratic equation [tex]\(y = at^2 + bt + c\)[/tex], occurs at the vertex. The time [tex]\(t\)[/tex] at which the vertex occurs can be found using the formula [tex]\(t = -\frac{b}{2a}\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are coefficients from the equation [tex]\(y = at^2 + bt + c\)[/tex].
Here, [tex]\(a = -16\)[/tex] and [tex]\(b = 46\)[/tex].
[tex]\[ t = -\frac{b}{2a} = -\frac{46}{2 \times -16} = \frac{46}{32} = 1.4375 \, \text{seconds} \][/tex]
2. Calculate the maximum height:
Substitute [tex]\(t = 1.4375\)[/tex] seconds back into the height equation to find [tex]\(y\)[/tex]:
[tex]\[ y = 46(1.4375) - 16(1.4375)^2 \][/tex]
Evaluate the first term:
[tex]\[ 46 \times 1.4375 = 66.125 \][/tex]
Next, evaluate the second term:
[tex]\[ 16 \times (1.4375)^2 = 16 \times 2.06640625 = 33.0625 \][/tex]
Therefore, the height [tex]\(y\)[/tex] at [tex]\(t = 1.4375\)[/tex] seconds is:
[tex]\[ y = 66.125 - 33.0625 = 33.0625 \, \text{feet} \][/tex]
3. Round the result to the nearest whole number:
Thus, the maximum height attained by the ball, rounded to the nearest whole number, is:
[tex]\[ 33 \, \text{feet} \][/tex]
So, the maximum height attained by the ball is [tex]\( \boxed{33} \)[/tex] feet.
[tex]\[ y = 46t - 16t^2 \][/tex]
To find the maximum height, we need to determine the highest point of this parabolic trajectory. We can do this via the following steps:
1. Determine the time at which the maximum height occurs:
The maximum height of a parabola that opens downward, given by a quadratic equation [tex]\(y = at^2 + bt + c\)[/tex], occurs at the vertex. The time [tex]\(t\)[/tex] at which the vertex occurs can be found using the formula [tex]\(t = -\frac{b}{2a}\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are coefficients from the equation [tex]\(y = at^2 + bt + c\)[/tex].
Here, [tex]\(a = -16\)[/tex] and [tex]\(b = 46\)[/tex].
[tex]\[ t = -\frac{b}{2a} = -\frac{46}{2 \times -16} = \frac{46}{32} = 1.4375 \, \text{seconds} \][/tex]
2. Calculate the maximum height:
Substitute [tex]\(t = 1.4375\)[/tex] seconds back into the height equation to find [tex]\(y\)[/tex]:
[tex]\[ y = 46(1.4375) - 16(1.4375)^2 \][/tex]
Evaluate the first term:
[tex]\[ 46 \times 1.4375 = 66.125 \][/tex]
Next, evaluate the second term:
[tex]\[ 16 \times (1.4375)^2 = 16 \times 2.06640625 = 33.0625 \][/tex]
Therefore, the height [tex]\(y\)[/tex] at [tex]\(t = 1.4375\)[/tex] seconds is:
[tex]\[ y = 66.125 - 33.0625 = 33.0625 \, \text{feet} \][/tex]
3. Round the result to the nearest whole number:
Thus, the maximum height attained by the ball, rounded to the nearest whole number, is:
[tex]\[ 33 \, \text{feet} \][/tex]
So, the maximum height attained by the ball is [tex]\( \boxed{33} \)[/tex] feet.
