An airplane flying horizontally 1000 meters above the ground at 200 km per hr drops a bomb on a target on the ground. Determine the acute angle between the vertical and the line joning the airplaod and target at the instant when the bomb is released.
t equals the fall time
0 = -4.9 t² + 1000
horizontal distance to target
d = t * 200 km/hr
convert km/hr to m/s
the tangent of the angle is the height of the plane, divided by the horizontal distance to target
To determine the acute angle between the vertical and the line joining the airplane and the bomb at the instant the bomb is released, we can use trigonometry.
Let's define the following variables:
- h: height of the airplane above the ground (1000 meters)
- d: horizontal distance between the airplane and the target (varies with time)
- θ: acute angle between the vertical and the line joining the airplane and the bomb
At the instant the bomb is released, the airplane and the bomb are at the same horizontal position. Therefore, the horizontal distance between the airplane and the target (d) is equal to the ground speed of the airplane multiplied by the time it takes for the bomb to reach the ground.
We need to convert the given ground speed of the airplane from km/hr to m/s:
200 km/hr * (1000 m / 1 km) * (1 hr / 3600 s) = 55.56 m/s
Let t be the time it takes for the bomb to reach the ground. Using the equation of motion for vertical motion under gravity, we can calculate the time as follows:
h = (1/2) * g * t^2
1000 = (1/2) * 9.8 * t^2
t^2 = 1000 / (0.5 * 9.8)
t^2 = 204.08
t ≈ √204.08
t ≈ 14.28 seconds
Now, we can calculate the horizontal distance (d) using the equation d = v * t, where v is the ground speed of the airplane:
d = 55.56 m/s * 14.28 s
d ≈ 793.848 meters
Using the given height (h) and the horizontal distance (d), we can calculate the tangent of the angle (θ) using the equation tan(θ) = h / d:
tan(θ) = 1000 / 793.848
θ ≈ tan^(-1)(1000 / 793.848)
Using a calculator, we find that:
θ ≈ 51.6 degrees
Therefore, the acute angle between the vertical and the line joining the airplane and the bomb at the instant when the bomb is released is approximately 51.6 degrees.
To determine the acute angle between the vertical and the line joining the airplane and the target when the bomb is released, we can use trigonometry.
First, let's draw a diagram to better understand the situation:
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Target| \
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|________\ Airplane
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\ Height|
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\ |
\ |
\ |
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In the diagram, the vertical direction is represented by a straight line, and the line connecting the airplane and the target is represented by a slanted line.
Now, let's calculate the acute angle using trigonometry.
The horizontal speed of the airplane is given as 200 km/hr. We need to convert it to meters per second since the other measurements are in meters. To convert km/hr to m/s, we divide by 3.6:
Speed of airplane = 200 km/hr / 3.6 = 55.56 m/s
We can consider the time it takes for the bomb to fall as 't' seconds. The horizontal distance (d) traveled by the airplane in 't' seconds can be calculated using the speed of the airplane:
d = Speed of airplane * t
Now, let's calculate the time it takes for the bomb to fall to the ground. This can be done using the equation of motion for free fall:
Height = (1/2) * acceleration due to gravity * t^2
Since the initial vertical velocity is 0 (the bomb is released from rest), we can write:
1000 = (1/2) * 9.8 * t^2
Simplifying the equation gives:
t^2 = 1000 / (0.5 * 9.8)
t^2 = 1000 / 4.9
t = sqrt(200) ≈ 14.14 seconds
Now, we can calculate the horizontal distance traveled by the airplane in this time:
d = Speed of airplane * t
d = 55.56 m/s * 14.14 s
d ≈ 785.74 meters
Therefore, the horizontal distance traveled by the airplane is approximately 785.74 meters.
Now, we have a right-angled triangle with the vertical height of 1000 meters and the horizontal distance of approximately 785.74 meters. The angle between the vertical and the line joining the airplane and the target is given by the inverse tangent of the vertical height divided by the horizontal distance:
Angle = InvTan(Vertical height / Horizontal distance)
Angle = InvTan(1000 / 785.74)
Angle ≈ 51.30 degrees
Hence, the acute angle between the vertical and the line joining the airplane and the target when the bomb is released is approximately 51.30 degrees.