Final answer:
The crate will hit the ground approximately 18.06 seconds after being dropped from the airplane and approximately 1232.652 meters in front of the release point.
Step-by-step explanation:
In order to find how far in front of the release point the crate will hit the ground, we need to consider the horizontal motion of the airplane and the vertical motion of the crate. The horizontal motion of the airplane does not affect the vertical motion of the crate because they are independent of each other. The vertical motion of the crate can be analyzed using equations of motion under constant acceleration due to gravity.
Since the crate is dropped from rest, its initial velocity in the vertical direction is zero. The final position of the crate is given as 1600 m above the ocean. We can use the equation:
y = yo + vot + 0.5gt^2
where y represents the final position (1600 m above the ocean), yo represents the initial position (0 m above the ocean), vo represents the initial velocity (zero), g represents the acceleration due to gravity (-9.8 m/s^2 for downward direction), and t represents the time taken.
Applying the values to the equation:
1600 m = 0 m + 0 m/s * t + 0.5 * (-9.8 m/s^2) * t^2
Simplifying and rearranging the equation, we get:
4.9t^2 = 1600 m
Dividing both sides by 4.9, we have:
t^2 = 326.53 s^2
Taking the square root of both sides, we find:
t = 18.06 s
Therefore, it will take approximately 18.06 seconds for the crate to hit the ground. Since the horizontal motion of the airplane and the vertical motion of the crate are independent, the distance traveled by the airplane horizontally during this time can be calculated using the equation:
d = v * t
where d represents the horizontal distance traveled, v represents the horizontal velocity of the airplane, and t represents the time taken (18.06 s).
Substituting the given values:
d = 68.2 m/s * 18.06 s
Simplifying the equation:
d = 1232.652 m
Therefore, the crate will hit the ground approximately 1232.652 meters in front of the release point.