A bomber on a military mission is flying horizontally at a weight of 3000m above the ground at 60km mn–¹,it drops a bomb on a target on the ground.determine the acute angle between the vertical and the line joining the bomber and the target at the instant the bomb is released.
60km mn–¹,?? what in the world is that. I assume you mean some speed, lets call it V.
Next, "weight"? is this supposed to be height? CHECK YOUR POSTS.
time to fall 30OOm: h=1/2 g t^2 or
time= sqrt (6000/9.8)=24.7 seconds
horizontal distance traveled in flight: V*t= 24.7 V
acute angle: arctan(3000/(24.7V))
To determine the acute angle between the vertical and the line joining the bomber and the target when the bomb is released, we can use trigonometry.
Let's analyze the situation:
1. The bomber is flying horizontally, meaning its flight path is parallel to the ground.
2. The bomber is flying at a height of 3000m above the ground, and its speed is 60 km/min.
3. The bomb is dropped from the bomber onto a target on the ground.
To find the angle, let's consider a right-angled triangle formed by the bomber, the target, and a vertical line from the bomber to the target.
We can use the tangent function to determine the angle. The tangent of an angle is equal to the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle.
In this case, the side opposite the angle is the height of the bomber (3000m), and the side adjacent to the angle is the horizontal distance the bomber travels in one minute (60 km/min).
Let's calculate the tangent of the angle:
Tangent(angle) = (height of bomber) / (horizontal distance traveled in one minute)
Tangent(angle) = 3000m / (60 km/min)
First, let's convert km/min to m/min, so the units are consistent:
60 km/min = 60,000 m/min
Tangent(angle) = 3000m / 60,000m
Simplifying the fraction:
Tangent(angle) = 1/20
Now, we can find the angle by finding the inverse tangent (arctan) of the tangent value using a calculator:
Angle = arctan(1/20)
Calculating this value, we get:
Angle ≈ 2.86 degrees
Therefore, the acute angle between the vertical and the line joining the bomber and the target at the instant the bomb is released is approximately 2.86 degrees.
To determine the acute angle between the vertical and the line joining the bomber and the target at the instant the bomb is released, we can make use of trigonometry.
The key information we need is the horizontal distance between the bomber and the target when the bomb is released.
Given that the bomber is flying horizontally at a height of 3000m above the ground, we can consider the height as the opposite side of a right triangle.
Next, we need to find the horizontal distance, which is the adjacent side of the right triangle. We can calculate this distance using the formula:
distance = speed × time
The bomber is flying at a speed of 60 km/h, which can be converted to meters per second by dividing by 3.6 (since 1 km/h = 1/3.6 m/s).
Thus, the speed of the bomber is:
60 km/h ÷ 3.6 = 16.67 m/s
We are given that the bomb is dropped, meaning the time of flight is zero. Therefore, the horizontal distance traveled by the bomber is also zero at the instant the bomb is released.
Now that we have the opposite side (height) as 3000m and the adjacent side (distance) as 0m, we can use the tangent function to find the acute angle:
tan(angle) = opposite / adjacent
tan(angle) = 3000 / 0
At this point, we have an undefined value for the tangent of the angle because we cannot divide by zero.
Since the horizontal distance is zero, it means the bomb is directly below the bomber at the instant of release. In other words, the line joining the bomber and the target is vertical, making the angle between the vertical and the line joining them 90 degrees.
Therefore, the acute angle between the vertical and the line joining the bomber and the target at the instant the bomb is released is 90 degrees.