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A gannet is a seabird that fishes by diving from a great height

If a gannet hits the water at 32 m/s , what height did it dive from? Assume that the gannet was motionless before starting its dive.

Respuesta :

egenriether
For constant acceleration along a given direction, we can relate acceleration, velocity and position with the following equation that doesn't involve time:
[tex] v^{2}= v_{0}^{2}+2a(x- x_{0}) [/tex]
In this equation x is the final position, which we take to be 0.  Also the initial velocity Vo is zero.  Thus the equation simplifies to 
[tex] v^{2}= 2a(- x_{0})[/tex]
Putting in v=32m/s, a=-9.81m/s^2 gives
[tex]32^{2}= 2(-9.81)(- x_{0}) [/tex]
[tex] \frac{32^{2}}{2*9.81}= x_{0}=52.19m [/tex]

The height that should be dive from is 52.24m.

Given that,

  • The gannet that should hit the water should be 32 m/s.

The following formula should be used for determining the height is:

[tex]h = \frac{v^2}{2g} \\\\= \frac{32}{2\times 9.8} \\\\[/tex]

= 52.24 m

Here

v denotes the speed.

And, g denotes the gravitational acceleration.

Therefore we can conclude that The height that should be dive from is 52.24m.

Learn more about the speed here: brainly.com/question/22610586

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