To find all the values of [tex]\( t \)[/tex] for which the ball's height is 10 feet, we start with the given quadratic equation for the ball's height:
[tex]\[ h(t) = 4 + 21t - 16t^2 \][/tex]
We need to find [tex]\( t \)[/tex] when the height [tex]\( h \)[/tex] is 10 feet.
[tex]\[ 4 + 21t - 16t^2 = 10 \][/tex]
Next, let's rearrange the equation to set it to zero:
[tex]\[ 4 + 21t - 16t^2 - 10 = 0 \][/tex]
Simplify the expression:
[tex]\[ -16t^2 + 21t - 6 = 0 \][/tex]
This is a quadratic equation in the standard form [tex]\( at^2 + bt + c = 0 \)[/tex], where:
[tex]\[ a = -16 \][/tex]
[tex]\[ b = 21 \][/tex]
[tex]\[ c = -6 \][/tex]
To solve this quadratic equation, we can use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ t = \frac{-21 \pm \sqrt{21^2 - 4(-16)(-6)}}{2(-16)} \][/tex]
[tex]\[ t = \frac{-21 \pm \sqrt{441 - 384}}{-32} \][/tex]
[tex]\[ t = \frac{-21 \pm \sqrt{57}}{-32} \][/tex]
We now calculate the two possible solutions for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-21 + \sqrt{57}}{-32} \][/tex]
[tex]\[ t = \frac{-21 - \sqrt{57}}{-32} \][/tex]
By evaluating these expressions, we get the two values of [tex]\( t \)[/tex]:
[tex]\[ t \approx 0.42 \][/tex]
[tex]\[ t \approx 0.89 \][/tex]
Thus, the values of [tex]\( t \)[/tex] for which the ball's height is 10 feet are:
[tex]\[ t = 0.42 \text{ seconds or } 0.89 \text{ seconds} \][/tex]
So the answer is:
[tex]\[ t = 0.42 \text{ or } 0.89 \text{ seconds} \][/tex]
