A hot-air balloon is rising straight up with a speed of 6 m/s. A ballast bag is released from rest relative to the balloon at 7 m above the ground. How much time elapses before the ballast bag hits the ground?
To determine the time it takes for the ballast bag to hit the ground, we can use kinematic equations. Specifically, we can use the equation that relates the displacement, initial velocity, time, and acceleration:
\[s = ut + \frac{1}{2}at^2\]
Where:
- s is the displacement (height)
- u is the initial velocity (6 m/s for the balloon)
- t is the time
- a is the acceleration due to gravity (-9.8 m/s^2 in the downward direction)
Given that the ballast bag is released from rest relative to the balloon at 7 m above the ground, the displacement (s) would be -7 m. We use a negative value because the displacement is in the opposite direction of the balloon's upward motion.
Substituting these values into the equation, we have:
\[-7 = 6t + \frac{1}{2}(-9.8)t^2\]
Rearranging the equation and dividing by -9.8, we get a quadratic equation:
\[4.9t^2 - 6t - 7 = 0\]
To solve for t, we can use the quadratic formula:
\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In this equation, a = 4.9, b = -6, and c = -7.
Calculating the values under the square root:
\[b^2 - 4ac = (-6)^2 - 4(4.9)(-7) = 36 + 137.2 = 173.2\]
Taking the square root of 173.2:
\[\sqrt{173.2} \approx 13.15\]
Using the quadratic formula:
\[t = \frac{-(-6) \pm 13.15}{2(4.9)}\]
Simplifying:
\[t = \frac{6 \pm 13.15}{9.8}\]
There are two possible solutions, one with the plus sign and one with the minus sign.
For the positive solution:
\[t = \frac{6 + 13.15}{9.8} \approx 1.98\]
So, the ballast bag takes approximately 1.98 seconds to hit the ground after it is released from the balloon.