Question

The equation $y = -6t^2 - 10t + 56$ describes the height (in feet) of a ball thrown downward at 10 feet per second from a height of 56 feet from the surface from Mars. In how many seconds will the ball hit the ground? Express your answer as a decimal rounded to the nearest hundredth.

Answer 1

2.33

Explanation:

Setting $y$ to zero, we find the following:

\begin{align*}

-6t^2 - 10t + 56 &= 0 \\

\Rightarrow \quad 6t^2 + 10t - 56 &= 0 \\

\Rightarrow \quad 3t^2 + 5t - 28 &= 0 \\

\Rightarrow \quad (3t-7)(t+4) &= 0.

\end{align*}As $t$ must be positive, we can see that $t = \frac{7}{3} \approx \boxed{2.33}.$

Answer 2

2.33 Seconds

Explanation:

The equation y = -6t^2 - 10t + 56 expresses the height of a ball at a given time, t, on the planet Mars. We are also given that the ball is thrown with a velocity of 10 feet per second with the initial height of 56 feet.

To find the amount of time to reach the ground, we can say that the time being found will be when the ball is on the ground, or when y = 0. So we simply set our equation to 0 and solve for t.

y = -6t^2 - 10t + 56

0 = -6t^2 - 10t + 56

0 = -1 (6t^2 + 10t + -56)

0 = -1 (3t - 7) (2t + 8)

(3t - 7) = 0 OR (2t + 8) = 0

3t = 7 OR 2t = -8

t = 7/3 OR t = -4

Since time will not be negative, we will want to choose the positive solution for this quadratic equation.

Hence, the amount of time for the ball to hit the ground will be 7/3 seconds or 2.33 seconds.

Cheers.