Respuesta :
Let the key is free falling, therefore from equation of motion
[tex]h = ut +\frac{1}{2}gt^2.[/tex].
Take initial velocity, u=0, so
[tex]h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2[/tex].
[tex]h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2 \\\ t =\sqrt{\frac{2h}{g} }[/tex]
As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula
[tex]d= v \times t[/tex]
From above substituting t,
[tex]d = v \times \sqrt{\frac{2h}{g} }[/tex].
Now substituting all the given values and g = 9.8 m/s^2, we get
[tex]d = 3.5 \ m/s \times \sqrt{\frac{2 \times 45 m}{9.8 m/s^2} } = 10.60 m[/tex].
Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.
