A hot-air balloon is rising upward with a constant speed of 2.10m/s. When the balloon is 3.18m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?
just solve for t in
3.18 + 2.1t - 4.9t^2 = 0
Pretty clumsy balloonist. Can't hold onto his compass for much over a second!
To find the time it takes for the compass to hit the ground, we can use the equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
where:
- \( s \) is the distance traveled (3.18m, the height of the balloon above the ground)
- \( u \) is the initial velocity (0m/s since the compass is dropped)
- \( t \) is the time we want to find
- \( a \) is the acceleration (acceleration due to gravity, which is approximately \( 9.8 \, \text{m/s}^2 \))
However, since the balloon is moving upward with a constant speed, the initial velocity of the compass is the same as the velocity of the balloon (2.10m/s). So we can rewrite the equation as:
\[ s = (2.10 \, \text{m/s}) t - \frac{1}{2} (9.8 \, \text{m/s}^2) t^2 \]
Rearranging the equation, we get:
\[ \frac{1}{2} (9.8 \, \text{m/s}^2) t^2 - (2.10 \, \text{m/s}) t + 3.18 \, \text{m} = 0 \]
This is a quadratic equation in the form \( at^2 + bt + c = 0 \), where \( a = \frac{1}{2} (9.8 \, \text{m/s}^2) \), \( b = -(2.10 \, \text{m/s}) \), and \( c = 3.18 \, \text{m} \).
We can solve this quadratic equation for \( t \) using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values, we get:
\[ t = \frac{-(-(2.10 \, \text{m/s})) \pm \sqrt{(-(2.10 \, \text{m/s})))^2 - 4(\frac{1}{2} (9.8 \, \text{m/s}^2))(3.18 \, \text{m})}}{2(\frac{1}{2} (9.8 \, \text{m/s}^2))} \]
Simplifying the equation further will give us the value of \( t \).