To find the time at which the ball reaches its maximum height and the maximum height itself, we need to analyze the quadratic function describing the height \( h \):
\[ h = -16t^2 + 36t + 9 \]
A quadratic function \( ax^2 + bx + c \) reaches its maximum (or minimum) value at \( t = \frac{-b}{2a} \). For our function:
- \( a = -16 \)
- \( b = 36 \)
So, the time at which the ball reaches maximum height is:
\[ t = \frac{-36}{2 \times (-16)} = \frac{-36}{-32} = 1.125 \text{ seconds} \]
Rounded to the nearest hundredth, it is \( 1.13 \) seconds.
Now, to find the maximum height, we substitute \( t = 1.125 \) back into the height function:
\[ h = -16(1.125)^2 + 36(1.125) + 9 \]
First, calculate \( (1.125)^2 \):
\[ (1.125)^2 = 1.265625 \]
Then, calculate:
\[ -16 \times 1.265625 = -20.25 \]
\[ 36 \times 1.125 = 40.5 \]
So,
\[ h = -20.25 + 40.5 + 9 \]
\[ h = 29.25 \text{ feet} \]
Thus, the ball reaches its maximum height of 29.25 feet at approximately 1.13 seconds. The correct answer is:
\[ \text{1.13 s; 29.25 ft} \]
1.
A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 36t + 9. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?
2.25 s; 9 ft
1.13 s; 69.75 ft
1.13 s; 29.25 ft
1.13 s; 31.5 ft
1 answer