To solve the problem, we will analyze and solve the given quadratic equation for the height [tex]\( h \)[/tex] in terms of time [tex]\( t \)[/tex]:
[tex]\[ h(t) = -4.9t^2 + 8.8t + 3 \][/tex]
### Part (a): Finding the time when Sam reaches the water
Sam reaches the water when his height [tex]\( h \)[/tex] is 0. Therefore, we need to solve the equation:
[tex]\[ -4.9t^2 + 8.8t + 3 = 0 \][/tex]
This is a quadratic equation in the form [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -4.9 \)[/tex], [tex]\( b = 8.8 \)[/tex], and [tex]\( c = 3 \)[/tex].
Using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], we substitute in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ t = \frac{-8.8 \pm \sqrt{8.8^2 - 4(-4.9)(3)}}{2(-4.9)} \][/tex]
First, calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 8.8^2 - 4(-4.9)(3) \][/tex]
[tex]\[ \Delta = 77.44 + 58.8 \][/tex]
[tex]\[ \Delta = 136.24 \][/tex]
Now, calculate the roots:
[tex]\[ t = \frac{-8.8 \pm \sqrt{136.24}}{-9.8} \][/tex]
Since the square root of 136.24 is approximately 11.67, we have:
[tex]\[ t = \frac{-8.8 \pm 11.67}{-9.8} \][/tex]
This gives us two possible solutions:
[tex]\[ t_1 = \frac{-8.8 + 11.67}{-9.8} = \frac{2.87}{-9.8} \approx -0.29 \][/tex]
[tex]\[ t_2 = \frac{-8.8 - 11.67}{-9.8} = \frac{-20.47}{-9.8} \approx 2.09 \][/tex]
Since time cannot be negative, we discard [tex]\( t_1 \)[/tex] and take [tex]\( t_2 \)[/tex]:
[tex]\[ t \approx 2.09 \][/tex]
Rounding to the nearest tenth:
[tex]\[ t \approx 2.1 \][/tex]
So, Sam is in the air for approximately 2.1 seconds before he reaches the water.
### Part (b): Finding the maximum height Sam reaches
To find the maximum height, we need to determine the vertex of the parabolic equation [tex]\( h(t) = -4.9t^2 + 8.8t + 3 \)[/tex]. The time at which the maximum height occurs can be found using the vertex formula [tex]\( t = -\frac{b}{2a} \)[/tex]:
[tex]\[ t = -\frac{8.8}{2(-4.9)} = \frac{8.8}{9.8} = 0.9 \][/tex]
Now, substitute [tex]\( t = 0.9 \)[/tex] back into the height equation to find the maximum height:
[tex]\[ h(0.9) = -4.9(0.9)^2 + 8.8(0.9) + 3 \][/tex]
[tex]\[ h(0.9) = -4.9(0.81) + 7.92 + 3 \][/tex]
[tex]\[ h(0.9) = -3.969 + 7.92 + 3 \][/tex]
[tex]\[ h(0.9) = 6.951 \][/tex]
Rounding to the nearest tenth:
[tex]\[ h(0.9) \approx 7.0 \][/tex]
So, the maximum height Sam reaches is approximately 7.0 meters.
### Summary
- Time in the air before reaching the water: approximately [tex]\( 2.1 \)[/tex] seconds
- Maximum height reached: approximately [tex]\( 7.0 \)[/tex] meters
