# 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 1, 2015

$39.62 \text{ m}$

#### Explanation:

Consider the Cartesian frame reference. Meaning that the downward direction is taken to be negative and the upward direction is positive $H$ stands for the maximum height
$h$ is the height at any time $t$ during the balloon's fall

From newton's equation of motion for displacement:
$y = u t + \frac{1}{2} a {t}^{2}$

In the present situation,

$y$ is the displacement
Considering the fact that, height $h$ at any time can be gotten from the equation: $h = H + y$

Thus, $y = h - H$

$t$ is time of fall,
$t = 2.4 - 1 = 1.4$

$a$ is acceleration
$a = - g = - 9.8 m {s}^{-} 2$

$u$ is initial velocity or velocity at time $t = 0$
$u = - 21 m {s}^{-} 1$

The negative signs means that they are directed downwards!

Substituting all the variables into $y = u t + \frac{1}{2} a {t}^{2}$ , we get:

$\implies h - H = \left(- 21\right) \left(1.4\right) + \frac{1}{2} \left(- 9.8\right) {\left(1.4\right)}^{2} = - 39.004$

$\implies h = H - 39.004$

To find $H$ we consider the case where the balloon goes all the way down and hits the ground

This time,
$t = 2.4 s$ and $y = - H$
The other data remain the unchanged.

$- H = \left(- 21\right) \left(2.4\right) + \frac{1}{2} \left(- 9.8\right) {\left(2.4\right)}^{2} = - 78.624$

$\implies H = 78.624$

$\implies h = 78.624 - 39.004 = \textcolor{b l u e}{39.62 \text{ m}}$