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?

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Answer 1 · 2015-09-01T12:17:53

$39.62" m"$

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:
$y=ut+1/2at^2$

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
$a= -g= -9.8 ms^-2$

$u$ is initial velocity or velocity at time $t=0$
$u=-21ms^-1$

The negative signs means that they are directed downwards!

Substituting all the variables into $y=ut+1/2at^2$ , we get:

$=>h-H=(-21)(1.4)+1/2(-9.8)(1.4)^2=-39.004$

$=>h=H-39.004$

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

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

$-H=(-21)(2.4)+1/2(-9.8)(2.4)^2=-78.624$

$=>H=78.624$

$=>h=78.624-39.004=color(blue)(39.62" m")$