To determine the time at which the height of the projectile reaches 20 feet, we start with the given height equation for the projectile, which is:
[tex]\[ s = -16t^2 + 50t \][/tex]
We need to find the time [tex]\( t \)[/tex] when the height [tex]\( s \)[/tex] is 20 feet. So, we set [tex]\( s = 20 \)[/tex]:
[tex]\[ 20 = -16t^2 + 50t \][/tex]
Next, we rearrange this equation to form a standard quadratic equation. The standard form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex]. Substituting in our equation, we get:
[tex]\[ -16t^2 + 50t - 20 = 0 \][/tex]
Now we have a quadratic equation:
[tex]\[ -16t^2 + 50t - 20 = 0 \][/tex]
To solve for [tex]\( t \)[/tex], we can use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are the coefficients from the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex]. In our equation:
[tex]\[ a = -16, \][/tex]
[tex]\[ b = 50, \][/tex]
[tex]\[ c = -20 \][/tex]
First, we calculate the discriminant ([tex]\( \Delta \)[/tex]):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 50^2 - 4(-16)(-20) \][/tex]
[tex]\[ \Delta = 2500 - 1280 \][/tex]
[tex]\[ \Delta = 1220 \][/tex]
Now, we find the two possible values for [tex]\( t \)[/tex] using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ t = \frac{-50 \pm \sqrt{1220}}{2(-16)} \][/tex]
This yields two solutions:
[tex]\[ t_1 = \frac{-50 + \sqrt{1220}}{2(-16)} \approx 0.47 \][/tex]
[tex]\[ t_2 = \frac{-50 - \sqrt{1220}}{2(-16)} \approx 2.65 \][/tex]
Thus, the times when the projectile reaches a height of 20 feet are approximately [tex]\( t_1 = 0.47 \)[/tex] seconds and [tex]\( t_2 = 2.65 \)[/tex] seconds.
