Question #caa2d

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Question #caa2d

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Answer 1 · 2016-03-06T06:46:33

$= 494.51 m$ $=494.51m$ rounded to second place of decimal

Explanation:

Let the seagull be flying along the horizontal when it pooped.
Due to gravity poop starts falling in the downward vertical direction. Let the height of the seagull at that instant of time be $h$ $h$ and $m$ $m$ be mass of poop. Since motion of seagull and of poop are perpendicular to each other so these are independent of each other.

As we are required to find out height $h$ $h$ , therefore, we consider only the vertical direction. We also assume the following:
1. From the time it started to fall to the time it hits the ground, acceleration due to gravity $g = 9.81 m s^{- 2}$ $g=9.81ms^-2$ is constant.
2. We ignore drag due to atmosphere.

Initial Potential energy of the poop $= m g h$ $=mgh$
When it hits the ground it potential energy $= 0$ $=0$ , as here $h = 0$ $h=0$ .
Change in potential energy between the two positions $= m g h$ $=mgh$

We know that Kinetic energy $= \frac{1}{2} m v^{2}$ $=1/2mv^2$ ,
Initial velocity in the vertical is $0$ $0$ , Hence initial kinetic energy $= 0$ $=0$
Final kinetic energy $\frac{1}{2} m v_{Impact}^{2}$ $1/2mv_(text{Impact})^2$
Change in Kinetic energy $\frac{1}{2} m v_{Impact}^{2} = \frac{1}{2} m \cdot {98.5}^{2}$ $1/2mv_(text{Impact})^2=1/2mcdot98.5^2$

Due to the law of conservation of energy,
Change in potential energy $=$ $=$ Change in Kinetic energy
$m g h = \frac{1}{2} m \cdot {98.5}^{2}$ $mgh=1/2mcdot98.5^2$ , inserting value of $g$ $g$
$cancelm cdot 9.81cdoth=1/2 cancelmcdot98.5^2$
$h=1/2 cdot98.5^2/9.81=494.51m$ rounded to second place of decimal

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