To solve this problem, we need to find the maximum height of the ball, which occurs at the vertex of the parabola described by the quadratic function \( h(t) = -16t^2 + 24t + 7 \).
The vertex form of a quadratic function \( ax^2 + bx + c \) has its vertex at \( t = -\frac{b}{2a} \).
Here, \( a = -16 \) and \( b = 24 \). Thus,
\[ t = -\frac{24}{2(-16)} = -\frac{24}{-32} = \frac{24}{32} = \frac{3}{4} = 0.75 \text{ seconds} \]
To find the maximum height, we substitute \( t = 0.75 \) back into the original function:
\[ h(0.75) = -16(0.75)^2 + 24(0.75) + 7 \]
First, we calculate \( (0.75)^2 \):
\[ (0.75)^2 = 0.5625 \]
Next, we substitute and compute:
\[ h(0.75) = -16(0.5625) + 24(0.75) + 7 \]
\[ h(0.75) = -9 + 18 + 7 \]
\[ h(0.75) = 16 \]
So, the ball's maximum height is 16 feet.
Therefore, the ball reaches its maximum height of 16 feet at \( 0.75 \) seconds.
The correct answer is:
\[ \boxed{0.75 \text{ s}; 16 \text{ ft}} \]
2.
A ball is thrown into the air with an upward velocity of 24 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 24t + 7. 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?
0.75 s; 34 ft
1.5 s; 7 ft
0.75 s; 13 ft
0.75 s; 16 ft
1 answer
4 answers