# A man stands on a crane and throws a water balloon down at at 21 m/s. He finds that it takes 2.4s for the balloon to hit the ground. What is its height one second before it hits the ground?

Sep 2, 2015

We know that the initial velocity is $\text{21 m/s}$ and that $\Delta t = \text{2.4 s}$. We also know that $| g | = {\text{9.80655 m/s}}^{2}$. Let us assume that down is positive.

One second before it hits the ground implies $1.4$ seconds have passed. So, what we want is $\Delta y$ after $\text{1.4 s}$ with ${v}_{i} = \text{21 m/s}$.

What we don't know is the final height after $\text{1.4 s}$. However, we can figure out how far it would have traveled in the full time and how far it would have traveled in the partial time.

$\Delta {y}_{\text{full" = 1/2g t_"full"^2 + v_(iy)t_"full}}$

$= \frac{1}{2} \left(9.80665\right) {\left(2.4\right)}^{2} \text{ m" + (21)(2.4) " m}$

$= \text{78.643 m}$

Now we can use that as the initial height and re-solve for the final height, after determining the partial fall distance.

$\Delta {y}_{\text{partial" = 1/2g t_"partial"^2 + v_(iy)t_"partial}}$

$= \frac{1}{2} \left(9.80665\right) {\left(1.4\right)}^{2} \text{ m" + (21)(1.4) " m}$

$= \text{39.010 m}$

Finally, treating the ground as $y = 0$:

$\Delta {y}_{\text{full" - Deltay_"partial}} = \left(\cancel{{y}_{i}} - 0\right) - \left(\cancel{{y}_{i}} - {y}_{f}\right)$

$= 78.643 - 39.010 = {y}_{f} = \textcolor{b l u e}{\text{39.633 m}}$