Answer
The suitcase will land a distance of 795 meters from the dog.
Work Step by Step
We can find $v_{0y}$ when the suitcase falls out of the plane.
$v_{0y} = v_0~sin(\theta) = (90.0~m/s)~sin(23.0^{\circ})$
$v_{0y} = 35.17~m/s$
We can find $v_y$ just before the suitcase hits the ground.
$v_y^2 = v_{0y}^2+2gy$
$v_y = \sqrt{v_{0y}^2+2gy}$
$v_y = \sqrt{(35.17)^2+(2)(-9.80~m/s^2)(-114~m)}$
$v_y = 58.92~m/s$ (toward the ground)
We can find the time $t$ it takes the suitcase to reach the ground.
$t = \frac{v_y-v_{0y}}{g} = \frac{(-58.92~m/s)-(35.17~m/s)}{-9.80~m/s^2}$
$t = ~9.60s$
We can find the horizontal range $x$ of the suitcase.
$x = v_x~t = (90.0~m/s)~cos(23.0^{\circ})(9.60~s)$
$x = 795~m$
The suitcase will land a distance of 795 meters from the dog.
