A rocket is fired from the ground and the speed is given by:
$$-4.9t^2 + 49t + 155$$ When does it hit (return to) the ground?
So first we did:
$$9.8t = 49 \implies t = 5$$
So it hits the ground in $10$ seconds. However the answer key says $12.53$.
After reviewing some of the answers and thinking about this, obviously the upward progress rate is going to be different from the downward one so we can't simply double the time we get from solving for the inflection point.
We would also have to know how far above the ground the rocket is and then solve for the time it takes to fall back to the ground.
So our conclusion is that the question was worded wrong, and the right wording is reflected in the accepted answer.
Best Answer
I think there's a mistake in your formulation of the problem. My guess is that the correct formulation is the following:
A rocket is launched and it is observed that the rocket's altitude, as a function of time, follows the equation $$ y(t) = -4.9 t^2 + 49t + 155, \qquad t>0 $$ When will the rocket hit the ground?
Solution: When the rocket hits the ground, the altitude is zero. So we simply need to solve the equation $$ y(t) = -4.9 t^2 + 49t + 155 = 0 $$ This is easy, as we can use the quadratic formula: $$ t = \frac{-49 \pm \sqrt{49^2 -4\times (-4.9) \times 155}}{2 \times (-4.9)} $$ since time is only positive, we can take the negative-signed solution, giving $t\approx 12.525$.