Question

An airplane is flying with a velocity of 99.0 m/s at an angle of 22.0 degree above the horizontal. When the plane is a distance 109 m directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment. How far from the dog with suitcase land? You can ignore air resistance.

Answer 1

The x and y componets of velocity of plane are,

$u_x= u \times \cos \theta= 99 \times \cos 22=91.79 ms^{-1}$

$u_y= u \times \sin \theta= 99 \times \sin 22= 37.086ms^{-1}$

Consider the motion of the bag along y axis,

$H= u_yt+\frac{1}{2}(-g)t^2$

$-109= 37.086t-\frac{1}{2}9.8t^2$

Solve for t,

using $t=\frac{ -b \pm \sqrt{b^2-4ac}}{2a}$

where $a=4.9$, $b=-37.086$ and $c=-109$

take the positive value of t, which is 9.83s

Now consider the horizontal motion of suit case,

There is no acceleration during horizontal motion,

Therefore $s= u_x \times t$

$s= 91.79 \times 9.83=902.3 m$

The suit case falls at a distance 902.3 m from dog.

Note:

The general sign convention says that all upward motion are considered to be positive and all downward motions are considered to be negative.

So, in this problem the suit case moves downwards, so $H=-109m$ and $g=-9.8ms^{-2}$