Zech tossed a water balloon into the air from the top of his shed in the backyard. Use GeoGebra to graph the situation using the formula y=−6t2+30t+8

. When will the balloon hit the ground? Round your answer to the nearest tenth of a second.(1 point)
The balloon will hit the ground at
seconds.

1 answer

To determine when the water balloon will hit the ground using the given equation \( y = -6t^2 + 30t + 8 \), we need to find the value of \( t \) when \( y = 0 \).

Setting the equation to zero: \[ -6t^2 + 30t + 8 = 0 \]

We can solve this quadratic equation using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our case, the coefficients are:

  • \( a = -6 \)
  • \( b = 30 \)
  • \( c = 8 \)

Plugging in these values into the quadratic formula gives: \[ t = \frac{-30 \pm \sqrt{30^2 - 4(-6)(8)}}{2(-6)} \] \[ t = \frac{-30 \pm \sqrt{900 + 192}}{-12} \] \[ t = \frac{-30 \pm \sqrt{1092}}{-12} \]

Calculating \( \sqrt{1092} \): \[ \sqrt{1092} \approx 33.0 \]

So, continuing with our calculation: \[ t = \frac{-30 \pm 33.0}{-12} \]

Now calculating the two possible values for \( t \):

  1. \( t = \frac{-30 + 33.0}{-12} = \frac{3.0}{-12} = -0.25 \) (not valid, as time cannot be negative)
  2. \( t = \frac{-30 - 33.0}{-12} = \frac{-63.0}{-12} = 5.25 \)

Thus, rounding to the nearest tenth: \[ t \approx 5.3 \text{ seconds} \]

Therefore, the balloon will hit the ground at approximately 5.3 seconds.