Spymaster Paul, flying a constant 215 km/h horizontally in a low flying helicopter, wants to drop secret documents into his contact's open car which is traveling 155 km/h on a level highway 78.0 m below. At what angle ( to the horizontal) should the car be in his sights when the packet is released?
Please can someone send me the answer for this question? i'm confused.
i am confius too why nobody answer this question?
To find the angle at which Spymaster Paul should drop the secret documents into his contact's open car, we need to analyze the situation and use trigonometry.
Let's break down the given information:
- The helicopter is flying at a constant speed of 215 km/h horizontally.
- The contact's car is traveling at a speed of 155 km/h on a level highway.
- The vertical distance between the helicopter and the car is 78.0 m.
First, we need to convert the speeds from km/h to m/s to ensure consistent units. Let's do that:
Speed of the helicopter = 215 km/h
= (215 * 1000) / (60 * 60) m/s
≈ 59.722 m/s
Speed of the car = 155 km/h
= (155 * 1000) / (60 * 60) m/s
≈ 43.056 m/s
Now, let's consider the time it takes for the documents to fall from the helicopter to the car. During this time, both the helicopter and the car would have traveled a certain distance.
The horizontal distance covered by the helicopter is the same as the horizontal distance covered by the car, which is the time multiplied by their respective velocities.
Distance covered by the helicopter = Distance covered by the car
Horizontal velocity of helicopter * Time taken = Horizontal velocity of car * Time taken
Now, let's solve for the time taken:
59.722 * t = 43.056 * t
59.722 = 43.056
Since the time taken is the same for both, we can cancel out the "t" variable.
Therefore, the horizontal distances covered by both the helicopter and the car are equal.
To find the angle at which the car should be in Spymaster Paul's sights, we can create a right triangle using the horizontal and vertical distances.
Let θ represent the angle between the horizontal and the line connecting the helicopter and the car. In this case, we need to find the value of tan(θ).
tan(θ) = vertical distance / horizontal distance
= 78.0 m / (43.056 m/s * t)
Substituting the previously calculated value of t into the equation:
tan(θ) = 78.0 m / (43.056 m/s * 59.722)
≈ 0.319
Now, to find the angle θ, we can take the arctan of 0.319 using a calculator or trigonometric table:
θ ≈ arctan(0.319)
≈ 17.25 degrees
Therefore, Spymaster Paul should aim at an angle of approximately 17.25 degrees to the horizontal when dropping the secret documents into his contact's open car.
To determine the angle at which the car should be in Spymaster Paul's sights when the packet is released, we can use trigonometry.
Let's break down the given information:
1. The helicopter is flying horizontally at a constant speed of 215 km/h.
2. The car is traveling at a speed of 155 km/h.
3. The car is 78.0 m below the helicopter.
To find the angle, we need to consider the horizontal and vertical components of the motion.
Horizontal Component:
Both the helicopter and the car are moving horizontally. Since there is no vertical acceleration, the horizontal component of the helicopter's velocity will remain constant at 215 km/h.
Vertical Component:
The vertical component of the car's velocity is negligible since it is traveling on a level highway.
Now, let's calculate the time it takes for the packet to fall from the helicopter to the car:
The distance between the helicopter and the car is constant, regardless of their speeds. Therefore, we can consider the horizontal distance covered by the helicopter to be the same as the vertical distance covered by the packet.
Using the formula for distance, speed, and time: distance = speed x time
The time taken by the packet to fall 78.0 m can be calculated as:
78.0 m = 0 m/s (initial vertical velocity of the packet) x time + (1/2) x 9.8 m/s² (acceleration due to gravity) x time²
Simplifying the equation, we get:
78.0 m = (1/2) x 9.8 m/s² x time²
Solving for time, we find:
time² = (2 x 78.0 m) / 9.8 m/s²
time² = 15.9184 s²
time ≈ 3.9896 s (approximated to 4 decimal places)
Now, let's calculate the horizontal distance covered by the car during this time:
The horizontal distance covered by the car is given by the product of its velocity and the time:
horizontal distance = velocity x time
horizontal distance = 155 km/h x (3.9896 s/3600 s) (converting km/h to m/s)
Simplifying the equation, we find:
horizontal distance ≈ 43.439 meters
Finally, we can calculate the angle at which the car should be in Spymaster Paul's sights when the packet is released:
Using the tangent function: tan(theta) = opposite/adjacent
tan(theta) = 78.0 m / 43.439 m
Taking the inverse tangent (arctan) of both sides to isolate theta:
theta ≈ arctan(78.0 m / 43.439 m)
After calculating this, we find that the angle is approximately 60.91 degrees (rounded to two decimal places).
Therefore, Spymaster Paul should aim the car at an angle of approximately 60.91 degrees to the horizontal when releasing the packet.