Answer :
To find the time at which the rocket will hit the ground, we must solve the quadratic equation \( y = -16x^2 + 116x + 75 \) for \( x \) when \( y = 0 \). That is, we need to find the positive root(s) of the equation:
\[ -16x^2 + 116x + 75 = 0 \]
This is a quadratic equation in the standard form of \( ax^2 + bx + c = 0 \), where \( a = -16 \), \( b = 116 \), and \( c = 75 \).
To find the roots of the quadratic equation, we can apply the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Let's calculate the discriminant first, which is \( \Delta = b^2 - 4ac \):
\[ \Delta = 116^2 - 4(-16)(75) \]
\[ \Delta = 13456 + 4800 \]
\[ \Delta = 18256 \]
Now, we can use the quadratic formula:
\[ x = \frac{-116 \pm \sqrt{18256}}{-32} \]
Take the square root of the discriminant:
\[ \sqrt{18256} = 135.115 \]
So we have:
\[ x = \frac{-116 \pm 135.115}{-32} \]
For the positive time value (since a negative time value does not make physical sense in this context), we will use the plus sign in \( \pm \):
\[ x = \frac{-116 + 135.115}{-32} \]
\[ x = \frac{19.115}{-32} \]
\[ x \approx -0.597 \]
Since this is a physical problem and time can not be negative, we use the negative sign in \( \pm \) to get a positive time:
\[ x = \frac{-116 - 135.115}{-32} \]
\[ x = \frac{-251.115}{-32} \]
\[ x \approx 7.847 \]
Thus, the rocket will hit the ground approximately \( 7.85 \) seconds after launch, to the nearest hundredth of a second.
