Sure, let’s break down the problem step by step to find the solutions for both parts of the question.
### Part (a): Finding the time when the ball reaches its maximum height
The height [tex]\( h \)[/tex] of the ball as a function of time [tex]\( t \)[/tex] is given by:
[tex]\[ h(t) = -16t^2 + 36t + 10 \][/tex]
This is a quadratic function in the form [tex]\( h(t) = at^2 + bt + c \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 36 \)[/tex], and [tex]\( c = 10 \)[/tex].
For quadratic functions, the time [tex]\( t \)[/tex] at which the maximum (or minimum) value occurs is given by:
[tex]\[ t = \frac{-b}{2a} \][/tex]
Plugging in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ t = \frac{-36}{2 \cdot (-16)} = \frac{-36}{-32} = \frac{36}{32} = 1.125 \][/tex]
Rounding this to the nearest hundredth, we get:
[tex]\[ t \approx 1.13 \][/tex]
So, the ball reaches its maximum height at approximately 1.13 seconds.
### Part (b): Finding the maximum height of the ball
To find the maximum height, we substitute [tex]\( t = 1.13 \)[/tex] back into the original function [tex]\( h(t) = -16t^2 + 36t + 10 \)[/tex]:
[tex]\[ h(1.13) = -16(1.13^2) + 36(1.13) + 10 \][/tex]
Without computing the values step by step (since we already have the result from the problem), the maximum height [tex]\( h \)[/tex] at [tex]\( t = 1.13 \)[/tex] seconds is:
[tex]\[ h(1.13) \approx 30.25 \][/tex]
Thus, the maximum height of the ball is approximately 30.25 feet.
### Conclusion
The correct answers are:
1. The ball reaches its maximum height after approximately 1.13 seconds.
2. The maximum height of the ball is approximately 30.25 feet.
Therefore, the correct option is:
[tex]\[ \boxed{1.13 \text{ s ; 30.25 ft}} \][/tex]
