To determine how long after the ball is thrown it hits the ground, we need to find the time [tex]\( t \)[/tex] when the height [tex]\( h \)[/tex] is 0.
We are given the height equation:
[tex]\[ h = 70 - 10t - 5t^2 \][/tex]
To find when the ball hits the ground, set [tex]\( h \)[/tex] to 0:
[tex]\[ 0 = 70 - 10t - 5t^2 \][/tex]
We'll rewrite the equation in standard quadratic form [tex]\( at^2 + bt + c = 0 \)[/tex]:
[tex]\[ -5t^2 - 10t + 70 = 0 \][/tex]
Next, we solve the quadratic equation:
[tex]\[ -5t^2 - 10t + 70 = 0 \][/tex]
We can use the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where:
[tex]\[ a = -5, \quad b = -10, \quad c = 70 \][/tex]
Plugging in the values, we get:
[tex]\[ t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(-5)(70)}}{2(-5)} \][/tex]
[tex]\[ t = \frac{10 \pm \sqrt{100 + 1400}}{-10} \][/tex]
[tex]\[ t = \frac{10 \pm \sqrt{1500}}{-10} \][/tex]
[tex]\[ t = \frac{10 \pm \sqrt{25 \cdot 60}}{-10} \][/tex]
[tex]\[ t = \frac{10 \pm 5\sqrt{60}}{-10} \][/tex]
[tex]\[ t = -1 \pm \frac{\sqrt{60}}{2} \][/tex]
Taking the approximate value of [tex]\(\sqrt{60} \approx 7.75\)[/tex]:
[tex]\[ t = -1 + 3.87 \approx 2.87 \][/tex]
[tex]\[ t = -1 - 3.87 \approx -4.87 \][/tex]
Since time cannot be negative in this physical context, we discard the negative solution.
Therefore, the ball hits the ground [tex]\( 2.87 \)[/tex] seconds after it is thrown.
So, we have:
[tex]\[ t = 2.87 \][/tex]
If there is more than one valid value of time, we indicate them with 'or'. In this particular case:
[tex]\[ t = 2.87 \text{ seconds} \][/tex]
