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

$39.62 m$ $39.62" m"$

Consider the Cartesian frame reference. Meaning that the downward direction is taken to be negative and the upward direction is positive

$H$ $H$ stands for the maximum height
$h$ $h$ is the height at any time $t$ $t$ during the balloon's fall

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

In the present situation,

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

Thus, $y = h - H$ $y=h-H$

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

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

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

The negative signs means that they are directed downwards!

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

$\Rightarrow h - H = (- 21) (1.4) + \frac{1}{2} (- 9.8) {(1.4)}^{2} = - 39.004$ $=>h-H=(-21)(1.4)+1/2(-9.8)(1.4)^2=-39.004$

$\Rightarrow h = H - 39.004$ $=>h=H-39.004$

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

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

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

$\Rightarrow H = 78.624$ $=>H=78.624$

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