An aircraft flying at a horizontally height of 8000m above the ground at 200 metre per second drops a bomb on it's target on the ground. Find the angle between the vertical and the line joining the aircraft and the target immediately the bomb is released.
How long does it take the bomb to fall 8000m? 4.9t^2 = 8000, so t=40 s
In 40s, the bomb will travel 8000m.
What a coinkidink!
So, the angle is just 45°
To find the angle between the vertical and the line joining the aircraft and the target, we can use trigonometry.
Let's assume the distance between the aircraft and the target as 'd'.
Given:
Height of the aircraft (h) = 8000 m
Velocity of the aircraft (v) = 200 m/s
When the bomb is dropped, it will follow a parabolic path due to the horizontal velocity of the aircraft.
Using the equation of motion for vertical motion, we have:
h = ut + (1/2)gt^2
Since the initial vertical velocity (u) is zero and the acceleration due to gravity (g) is approximately 9.8 m/s^2, we can rewrite the equation as:
h = (1/2)gt^2
Solving for time (t), we have:
t^2 = (2h)/g
t = sqrt((2h)/g)
Substituting the given values, we get:
t = sqrt((2 * 8000)/9.8)
t ≈ 40.408 seconds
Now, we can find the horizontal distance traveled by the bomb during this time:
distance = velocity * time
distance = 200 * 40.408
distance ≈ 8081.6 meters
The angle between the vertical and the line joining the aircraft and the target can be found using the tangent function:
tan(theta) = (h/distance)
tan(theta) = (8000/8081.6)
Taking the inverse tangent of both sides, we find:
theta ≈ atan(8000/8081.6) ≈ 44.063 degrees
Therefore, the angle between the vertical and the line joining the aircraft and the target is approximately 44.063 degrees.
To find the angle between the vertical and the line joining the aircraft and the target, we need to consider the horizontal and vertical components of the aircraft's motion.
Let's break down the information given:
Height of the aircraft above the ground (h) = 8000m
Horizontal velocity of the aircraft (Vx) = 200 m/s
When the bomb is released, the horizontal motion of the aircraft doesn't affect the vertical motion. Therefore, we only need to focus on the vertical component.
In vertical motion, the initial velocity (u) is 0, as the bomb is dropped from rest.
We can use the equation of motion to calculate the time taken for the bomb to reach the ground:
h = ut + (1/2)gt^2
Since u = 0, the equation simplifies to:
h = (1/2)gt^2
Solving for t:
2h = gt^2
t^2 = (2h/g)
Now, we can find the time taken (t) for the bomb to reach the ground by taking the square root:
t = √(2h/g)
Substituting the given values:
t = √(2 * 8000 / 9.8)
t ≈ √(1632.65)
t ≈ 40.4 seconds
Now that we know the time taken for the bomb to reach the ground, we can calculate the vertical displacement (s) of the bomb using the equation:
s = ut + (1/2)gt^2
Since u = 0, the equation simplifies to:
s = (1/2)gt^2
Substituting the known values:
s = (1/2) * 9.8 * (40.4)^2
s ≈ 8000 meters
Now, we have the vertical displacement (s) and the horizontal displacement (distance traveled by the aircraft) is also given as 8000 meters.
To find the angle (θ) between the vertical and the line joining the aircraft and the target, we can use the tangent function:
tan(θ) = vertical displacement / horizontal displacement
tan(θ) = 8000 / 8000
tan(θ) = 1
Using inverse tangent (arctan), we find:
θ = arctan(1)
θ ≈ 45 degrees
Therefore, the angle between the vertical and the line joining the aircraft and the target immediately after the bomb's release is approximately 45 degrees.