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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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 · 2015-09-02T06:04:53

We know that the initial velocity is $"21 m/s"$ and that $Deltat = "2.4 s"$ . We also know that $|g| = "9.80655 m/s"^2$ . 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 $Deltay$ after $"1.4 s"$ with $v_i = "21 m/s"$ .

What we don't know is the final height after $"1.4 s"$ . 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.

$Deltay_"full" = 1/2g t_"full"^2 + v_(iy)t_"full"$

$= 1/2(9.80665)(2.4)^2 " m" + (21)(2.4) " m"$

$= "78.643 m"$

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

$Deltay_"partial" = 1/2g t_"partial"^2 + v_(iy)t_"partial"$

$= 1/2(9.80665)(1.4)^2 " m" + (21)(1.4) " m"$

$= "39.010 m"$

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

$Deltay_"full" - Deltay_"partial" = (cancel(y_i) - 0) - (cancel(y_i) - y_f)$

$= 78.643 - 39.010 = y_f = color(blue)("39.633 m")$

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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?

1 Answer

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