A millitary bomber is flies at a distance of 3000m,and above the ground 60km/min.determined the acute angle of the vertical and the line after it has been release
To determine the acute angle between the vertical and the line after the military bomber has been released, we need to use trigonometry. Let's assume that the line represents the bomber's flight path.
First, we need to calculate the time it takes for the bomber to cover the distance of 3000m at a speed of 60 km/min.
Since the speed of the bomber is given in km/min, let's convert the distance of 3000m to kilometers. We know that 1 kilometer is equal to 1000 meters, so 3000 meters is equal to 3 kilometers.
Next, we divide the distance by the speed to get the time it takes: 3 km / 60 km/min = 0.05 min.
Now, we can find the angle by considering the right triangle formed by the vertical, the distance covered by the bomber, and the ground.
The tangent of an angle in a right triangle is equal to the length of the side opposite the angle divided by the length of the side adjacent to the angle.
In this case, the side opposite the angle is the distance covered by the bomber (3 km) and the side adjacent to the angle is the height above the ground (60 km/min × 0.05 min = 3 km).
So, the tangent of the acute angle can be expressed as: tan(angle) = (3 km) / (3 km).
Now we can solve for the angle. Taking the inverse tangent (arctan) of both sides, we get:
angle = arctan[(3 km) / (3 km)].
Simplifying further, we find:
angle = arctan(1) ≈ 45°.
Therefore, the acute angle between the vertical and the line after the bomber has been released is approximately 45 degrees.