agent dave, flying a constant 215 km/hr horizontally in a low - flying helicopter wants to drop a small explosive onto a master criminal automobile travelling 155km/hr on a level highway 78m below. At what angle (with the horizontal) should the car be in his sights when the bomb is released
To determine the angle at which Agent Dave should release the bomb, we need to consider the relative motion of the helicopter and the car.
Given:
- The helicopter's horizontal speed: 215 km/hr
- The car's horizontal speed: 155 km/hr
- The height difference between the helicopter and the car: 78 m
First, let's convert the given speeds from km/hr to m/s for consistency:
Helicopter's speed: 215 km/hr = (215 * 1000) m/3600 s = 59.72 m/s
Car's speed: 155 km/hr = (155 * 1000) m/3600 s = 43.06 m/s
Now, let's determine the time it takes for the bomb to fall from the helicopter to the car. Since both the helicopter and the car are moving horizontally, the horizontal positions will remain unchanged during the fall. We only need to consider the vertical distance.
Using the formula s = ut + (1/2)at^2, where s is the vertical distance, u is the initial vertical velocity, t is the time, and a is the acceleration due to gravity, we can calculate the time.
Vertical distance (s) = 78 m (height difference)
Acceleration due to gravity (a) = 9.8 m/s^2 (assuming no other vertical forces)
Using the fact that the initial vertical velocity (u) is zero, and rearranging the formula, we have:
s = (1/2)at^2
2s = at^2
t^2 = (2s)/a
t = sqrt((2s)/a)
Substituting the values, we have:
t = sqrt((2 * 78) / 9.8) = sqrt(15.9) = 3.99 s (approximately)
Now, let's calculate the horizontal distance traveled by each vehicle during this time.
Distance traveled by the helicopter = (helicopter's speed) * (time)
Distance traveled by the car = (car's speed) * (time)
Distance traveled by the helicopter = 59.72 m/s * 3.99 s = 238.27 m
Distance traveled by the car = 43.06 m/s * 3.99 s = 171.72 m
The angle needed can be calculated using the tangent function:
Angle = arctan((vertical distance) / (horizontal distance))
Angle = arctan(78 m / (238.27 m - 171.72 m))
Angle = arctan(78 m / 66.55 m)
Angle ≈ 51.4°
Therefore, Agent Dave should aim the car at an angle of approximately 51.4° with the horizontal when releasing the bomb.
To determine the angle at which Agent Dave should aim the bomb, we can break down the problem into two components: the horizontal component and the vertical component.
First, let's calculate the time it takes for the bomb to reach the ground. We can use the equation:
Vertical distance (d) = 0.5 * acceleration due to gravity (g) * time squared (t^2)
Since the helicopter is flying 78 meters above the highway, the vertical distance is 78 meters. The acceleration due to gravity (g) is approximately 9.8 m/s². Plugging in these values, we get:
78 = 0.5 * 9.8 * t^2
Simplifying the equation:
t^2 = 78 / (0.5 * 9.8)
t^2 = 7.969
Taking the square root of both sides:
t = √7.969
t ≈ 2.82 seconds
Now, let's calculate the horizontal distance traveled by the car during this time. The horizontal distance is given by the equation:
Horizontal distance (d) = velocity (v) * time (t)
The velocity of the car is given as 155 km/hr, which we need to convert to m/s:
155 km/hr * (1000 m / 1 km) * (1 hr / 3600 s) = 43.06 m/s
Plugging in the values, we get:
d = 43.06 m/s * 2.82 s
d ≈ 121.32 meters
Now we have both the vertical distance and the horizontal distance. To find the angle, we can use the tangent function:
tan(θ) = vertical distance / horizontal distance
tan(θ) = 78 m / 121.32 m
θ = arctan(78 / 121.32)
θ ≈ 34.98 degrees
Therefore, Agent Dave should aim the car in his sights at approximately 34.98 degrees with the horizontal when releasing the bomb.