To find the values of [tex]\( t \)[/tex] where the height [tex]\( h \)[/tex] is 75 meters, given the initial velocity [tex]\( v \)[/tex] of 40 meters per second, we need to solve the equation given by the formula for height:
[tex]\[ h = v t - 5 t^2 \][/tex]
First, let's plug in the known values:
[tex]\[ h = 75 \][/tex]
[tex]\[ v = 40 \][/tex]
So the equation becomes:
[tex]\[ 75 = 40t - 5t^2 \][/tex]
Next, we rearrange the equation into standard quadratic form:
[tex]\[ 5t^2 - 40t + 75 = 0 \][/tex]
This is a quadratic equation in the form of [tex]\( ax^2 + bx + c = 0 \)[/tex]. Here, [tex]\( a = 5 \)[/tex], [tex]\( b = -40 \)[/tex], and [tex]\( c = 75 \)[/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 of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ t = \frac{-(-40) \pm \sqrt{(-40)^2 - 4 \cdot 5 \cdot 75}}{2 \cdot 5} \][/tex]
[tex]\[ t = \frac{40 \pm \sqrt{1600 - 1500}}{10} \][/tex]
[tex]\[ t = \frac{40 \pm \sqrt{100}}{10} \][/tex]
[tex]\[ t = \frac{40 \pm 10}{10} \][/tex]
This gives us two solutions:
[tex]\[ t_1 = \frac{40 + 10}{10} = \frac{50}{10} = 5 \][/tex]
[tex]\[ t_2 = \frac{40 - 10}{10} = \frac{30}{10} = 3 \][/tex]
So the two possible values for [tex]\( t \)[/tex] are [tex]\( 3 \)[/tex] and [tex]\( 5 \)[/tex] seconds.
We now look at our answer choices:
A. [tex]\( t = 1, t = 3 \)[/tex]
B. [tex]\( t = 1, t = 15 \)[/tex]
C. [tex]\( t = 3, t = 5 \)[/tex]
D. [tex]\( t = 5, t = 15 \)[/tex]
The correct answer is:
C. [tex]\( t = 3, t = 5 \)[/tex]
