Answer:
The pelican was travelling at approximately [tex]9.5\; {\rm m\cdot s^{-1}}[/tex] when it dropped the fish.
The fish hit the water at approximately [tex]16\; {\rm m\cdot s^{-1}}[/tex].
(Assumption: air resistance on the fish is negligible; [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)
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 [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] under gravity.
Since acceleration in the vertical direction is constant, make use of the SUVAT equation [tex]x = (1/2)\, a\, t^{2} + v_{0}\, t[/tex] to find the duration [tex]t[/tex] of the descent:
- The vertical position of the fish changed by [tex]x = 7.8\; {\rm m}[/tex] during the entire descent.
- The vertical acceleration of the fish is [tex]a = g = 9.81\; {\rm m\cdot s^{-2}}[/tex]
- The pelican was flying horizontally when it dropped the fish. Thus, the initial velocity of the fish in the vertical direction is [tex]v_{0} = 0\; {\rm m\cdot s^{-1}}[/tex]. The equation becomes [tex]x = (1/2)\, a\, t^{2}[/tex].
Rearrange this equation to find the duration of the descent, [tex]t[/tex]:
[tex]\begin{aligned} t &= \sqrt{\frac{2\, x}{a}} \\ &= \sqrt{\frac{2 \times 7.8\; {\rm m}}{9.81 \; {\rm m\cdot s^{-2}}} \\ &\approx 1.261\; {\rm s}\end{aligned}[/tex].
The vertical speed of the fish right before impact would be:
[tex]\begin{aligned} v_{y} &= a\, t \\ &\approx 9.81\; {\rm m\cdot s^{-2}} \times 1.261\; {\rm s} \\ &\approx 12.37\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].
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 [tex]x = 12.0 \; {\rm m}[/tex] within the [tex]t \approx 1.26\; {\rm s}[/tex] of the descent. As a result, the horizontal velocity of the fish would be:
[tex]\begin{aligned} v_{x} &= \frac{x}{t}\\ &\approx \frac{12.0\; {\rm m}}{1.261\; {\rm s}} \\ &\approx 9.52\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].
Hence, the initial speed of the pelican would be approximately [tex]9.5\; {\rm m\cdot s^{-1}}[/tex].
Thus, right before impact:
- The fish would be travelling at a horizontal velocity of [tex]v_{x} \approx 9.52\; {\rm m\cdot s^{-1}}[/tex].
- The fish would be travelling at a vertical velocity of [tex]v_{y} \approx 12.37\; {\rm m \cdot s^{-1}}[/tex].
Apply the Pythagorean Theorem to find the overall velocity of the fish at that moment:
[tex]\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}[/tex].
In other words, the fish would hit the water at approximately [tex]16\; {\rm m\cdot s^{-1}}[/tex].
