A bomber on a millitary mission is flying horizontally at a height of 3000m above the ground at 60km per minutes,it drops a bomb on a target on a ground. Determine the acute angle between the vertical and the line joining the bomber and the target at the instant the bomb is release.
how long is the bomb in the air?
0 = 3000 - (1/2) g t^2
t^2 = 6000/9.81
t = 24.7 seconds = 0.412
horizontal distance = 60,000 meters / 0.412 min
To determine the acute angle between the vertical and the line joining the bomber and the target, we can use trigonometry.
Let's break down the problem step by step:
1. We are given that the bomber is flying horizontally, which means its altitude is constant. Therefore, the angle between the vertical and the line joining the bomber and the target is the same as the angle between the line joining the bomber and the target and the horizontal.
2. We know the velocity of the bomber is 60 km per minute, but we need to convert it to meters per second to be consistent with the altitude given in meters.
60 km per minute = 60,000 meters per minute (since 1 km = 1,000 meters)
= 60,000 / 60 seconds per minute
= 1,000 meters per second
3. Now, we have the velocity of the bomber, and we can use it to find the time it takes for the bomb to reach the ground.
The formula to find the time is:
time = altitude / velocity
Using the given altitude, time = 3000 meters / 1000 meters per second = 3 seconds
4. During this time, the bomber covers a horizontal distance equal to its velocity multiplied by the time.
Distance = velocity * time = 1000 meters per second * 3 seconds = 3000 meters
5. Now, we have the horizontal distance (3000 meters), and the altitude (3000 meters) forms a right-angled triangle with the line joining the bomber and the target, which is the hypotenuse of the triangle.
Using trigonometry, we can find the acute angle opposite the altitude:
sin(angle) = altitude / hypotenuse
sin(angle) = 3000 meters / 3000 meters (since the hypotenuse is the same as the horizontal distance)
sin(angle) = 1
6. To find the angle, we take the inverse sine (or arcsine) of 1:
angle = arcsin(1) = 90°
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.
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 consider the situation. The bomber is flying horizontally at a height of 3000m above the ground. The distance from the bomber to the target is not given, but it is not necessary to find the angle.
To find the angle, we need to use the tangent function, which is defined as the ratio of the opposite side to the adjacent side in a right triangle.
In this case, the adjacent side is the height of the bomber above the ground (3000m), and the opposite side is the distance between the bomber and the target.
Let's assume the distance between the bomber and the target is 'd'.
We need to find the tangent of the angle, which is given by the formula:
tan(angle) = opposite/adjacent
tan(angle) = d/3000
Now, we can solve for the angle by taking the inverse tangent (arctan) of both sides of the equation:
angle = arctan(d/3000)
So, to find the acute angle, you need to know the distance between the bomber and the target (d). Once you have that value, you can substitute it into the equation above to obtain the acute angle.