An airplane is dropping bales of hay to cattle stranded in a blizzard on the Great Plains. The pilot releases the bales at 160 {\rm m} above the level ground when the plane is flying at 70.0 {\rm m}/{\rm s} 60.0^\circ above the horizontal.How far in front of the cattle should the pilot release the hay so that the bales will land at the point where the cattle are stranded?

d=632

To find how far in front of the cattle the pilot should release the hay, we need to determine the horizontal distance traveled by the bales.

First, we need to find the time it takes for the bales to reach the ground. We can use the vertical motion equation:

h = (1/2) * g * t²

Where:
h = vertical distance (160m)
g = acceleration due to gravity (9.8 m/s²)
t = time

Rearranging the equation to solve for t, we get:

t = sqrt((2h) / g)

Substituting the given values, we get:

t = sqrt((2 * 160) / 9.8) ≈ 6.38 seconds

Now, we can find the horizontal distance by multiplying the time by the horizontal component of the velocity. We use the equation:

d = v * t

Where:
d = distance
v = horizontal component of velocity (v * cosθ)
t = time

Substituting the given values, we get:

d = (70.0 * cos(60.0°)) * 6.38 ≈ 195.6 meters

Therefore, the pilot should release the hay approximately 195.6 meters in front of the cattle.