Question

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?

Answer 1

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")`