Answer
The pilot should release the hay a horizontal distance of 629 meters from the cattle.
Work Step by Step
We can find $v_{0y}$ when the hay leaves the plane.
$v_{0y} = v_0~sin(\theta) = (75~m/s)~sin(55^{\circ})$
$v_{0y} = 61.44~m/s$
We can find $v_y$ when the hay lands on the ground.
$v_y^2 = v_{0y}^2+2gy$
$v_y = \sqrt{v_{0y}^2+2gy}$
$v_y = \sqrt{(61.44~m/s)^2+(2)(-9.80~m/s^2)(-150~m)}$
$v_y = 81.94~m/s$ (toward the ground)
We can find the time $t$ for the hay's flight in the air.
$t = \frac{v_y-v_{0y}}{g} = \frac{(-81.94~m/s)-(61.44~m/s)}{-9.80~m/s^2}$
$t = 14.63~s$
We can find the horizontal distance $x$ that the hay travels.
$x = v_x~t = (75~m/s)~cos(55^{\circ})(14.63~s)$
$x = 629~m$
The pilot should release the hay a horizontal distance of 629 meters from the cattle.
