Question

A pelican flying along a horizontal path drops a fish from a height of 7.8 m. The fish travels 12.0 m horizontally before it hits the water below. What is the pelican’s initial speed? At what velocity does the fish hit the water?

Answer 1

The pelican was travelling at approximately
`9.5\; {\rm m\cdot s^(-1)}` when it dropped the fish.

The fish hit the water at approximately
`16\; {\rm m\cdot s^(-1)}`.

(Assumption: air resistance on the fish is negligible;
`g = 9.81\; {\rm m\cdot s^(-2)}`.)

Step-by-step explanation:

During the descent:

• In the horizontal direction, the fish travels at a constant velocity.
• In the vertical direction, the fish accelerates downwards at a constant rate of
  `g = 9.81\; {\rm m\cdot s^(-2)}` under gravity.

Since acceleration in the vertical direction is constant, make use of the SUVAT equation
`x = (1/2)\, a\, t^(2) + v_(0)\, t` to find the duration
`t` of the descent:

• The vertical position of the fish changed by
  `x = 7.8\; {\rm m}` during the entire descent.
• The vertical acceleration of the fish is
  `a = g = 9.81\; {\rm m\cdot s^(-2)}`
• The pelican was flying horizontally when it dropped the fish. Thus, the initial velocity of the fish in the vertical direction is
  `v_(0) = 0\; {\rm m\cdot s^(-1)}`. The equation becomes
  `x = (1/2)\, a\, t^(2)`.

Rearrange this equation to find the duration of the descent,
`t`:

`\begin{aligned} t &= \sqrt{(2\, x)/(a)} \\ &= \sqrt{\frac{2 * 7.8\; {\rm m}}{9.81 \; {\rm m\cdot s^(-2)}} \\ &\approx 1.261\; {\rm s}\end{aligned}`.

The vertical speed of the fish right before impact would be:

`\begin{aligned} v_(y) &= a\, t \\ &\approx 9.81\; {\rm m\cdot s^(-2)} * 1.261\; {\rm s} \\ &\approx 12.37\; {\rm m\cdot s^(-1)} \end{aligned}`.

In the horizontal direction, the speed of the fish was constant- same as the initial speed of the pelican. The fish travelled a horizontal distance of
`x = 12.0 \; {\rm m}` within the
`t \approx 1.26\; {\rm s}` of the descent. As a result, the horizontal velocity of the fish would be:

`\begin{aligned} v_(x) &= (x)/(t)\\ &\approx \frac{12.0\; {\rm m}}{1.261\; {\rm s}} \\ &\approx 9.52\; {\rm m\cdot s^(-1)} \end{aligned}`.

Hence, the initial speed of the pelican would be approximately
`9.5\; {\rm m\cdot s^(-1)}`.

Thus, right before impact:

• The fish would be travelling at a horizontal velocity of
  `v_(x) \approx 9.52\; {\rm m\cdot s^(-1)}`.
• The fish would be travelling at a vertical velocity of
  `v_(y) \approx 12.37\; {\rm m \cdot s^(-1)}`.

Apply the Pythagorean Theorem to find the overall velocity of the fish at that moment:

`\begin{aligned} v &= \sqrt{{v_(x)}^(2) + {v_(y)}^(2)} \\ &\approx \sqrt{{(9.52\; {\rm m\cdot s^(-1)})}^(2) + {(12.37\; {\rm m\cdot s^(-1)})}^(2)} \\ &\approx 16\; {\rm m\cdot s^(-1)}\end{aligned}`.

In other words, the fish would hit the water at approximately
`16\; {\rm m\cdot s^(-1)}`.