A plane diving with constant speed at an angle of 52.5° with the vertical, releases a package at an altitude of 499 m. The package hits the ground 3.87 s after release. How far horizontally does the package travel?
To find the horizontal distance the package traveled, we need to use the time of flight and the horizontal component of the velocity.
The given angle is between the plane's diving direction and the vertical. Since the plane is diving at an angle of 52.5°, the angle made with the horizontal is 90° - 52.5° = 37.5°.
Let's break down the problem into horizontal and vertical components:
1. Vertical Component:
The package is released at an altitude of 499 m. The vertical distance traveled by the package can be determined using the equation of motion: h = ut + (1/2)gt^2, where h is the vertical distance, u is the initial vertical velocity, g is the acceleration due to gravity (approx. 9.8 m/s^2), and t is the time of flight.
In this case, the final vertical distance (h) is 0 m (as the package hits the ground) and the initial vertical velocity (u) is 0 m/s (since the package is released from rest). Therefore, the equation simplifies to: 0 = (1/2)gt^2, which gives us t = sqrt((2h)/g).
Plugging in the given values, we get t = sqrt((2 * 499) / 9.8) ≈ 10.03 s. However, we are interested in the time the package takes to hit the ground after release, which is 3.87 s. Therefore, we can conclude that the package spends 3.87 s in the air before hitting the ground.
2. Horizontal Component:
The horizontal distance traveled by the package can be determined using the equation: d = vt, where d is the horizontal distance, v is the horizontal component of the velocity, and t is the time taken.
To find the horizontal component of the velocity, we can use trigonometry. In this case, the angle made with the horizontal is 37.5°. Thus, the horizontal component of the velocity (v) can be calculated by multiplying the total velocity (constant speed) by the cosine of the angle: v = total velocity * cos(angle).
We are not given the total velocity, but since it's a constant speed dive, the magnitude of the velocity will remain the same throughout. So, we can assume that the magnitude of the velocity is equal to the magnitude of the horizontal component of the velocity.
Therefore, we can calculate v by using the equation: v = (vertical velocity) / sin(angle).
To find the vertical velocity, we can again use trigonometry. The angle given with the vertical is 52.5°. So, the vertical component of the velocity (vertical velocity) can be calculated by multiplying the total velocity (constant speed) by the sine of the angle: vertical velocity = total velocity * sin(angle).
Now, we can substitute the given values and solve for v:
vertical velocity = total velocity * sin(52.5°),
horizontal component of the velocity (v) = vertical velocity / sin(37.5°).
Since we don't know the total velocity, we can assign it as a variable 'V' for convenience.
Now, substituting the given values and using the equation v = (vertical velocity) / sin(37.5°):
(vertical velocity) = V * sin(52.5°).
horizontal component of the velocity (v) = V * sin(52.5°) / sin(37.5°).
Finally, substituting the value of the time (t) of 3.87 s, we can calculate the horizontal distance (d) using the equation:
d = v * t.
Therefore, to find the horizontal distance the package traveled, we need to solve the equation:
d = (V * sin(52.5°) / sin(37.5°)) * 3.87 s.
This equation can be solved using a calculator to obtain the result.
To find the horizontal distance traveled by the package, we can use the formula for horizontal distance:
horizontal distance = (horizontal speed) x (time)
Since the plane is diving with a constant speed, the horizontal speed of the package will remain the same throughout its flight. However, we need to find the horizontal speed first.
We can use the trigonometric relationship for the vertical and horizontal components of motion:
vertical velocity = (initial vertical speed) + (acceleration x time)
initial vertical speed = (initial speed) x sin(θ)
acceleration = 9.8 m/s^2 (acceleration due to gravity)
Given:
angle of 52.5° with the vertical (θ = 52.5°)
altitude of 499 m
time of flight of 3.87 s
First, let's find the initial vertical speed:
initial vertical speed = (initial speed) x sin(θ)
Since the plane is diving with a constant speed, the initial speed can also be considered the constant horizontal speed.
Now we can plug in the given values:
θ = 52.5°
altitude = 499 m
Using a bit of trigonometry:
sin(θ) = opposite/hypotenuse
sin(θ) = altitude/initial speed
Rearranging the formula:
initial speed = altitude/sin(θ)
Plugging in the values:
initial speed = 499 m / sin(52.5°)
Now, let's calculate the initial speed:
initial speed = 499 m / 0.7855 (rounded to 4 decimal places)
initial speed ≈ 635.5227 m/s
Now that we have the initial speed, we can calculate the horizontal speed:
horizontal speed = initial speed x cos(θ)
Plugging in the values:
horizontal speed = 635.5227 m/s x cos(52.5°)
Now let's calculate the horizontal speed:
horizontal speed = 635.5227 m/s x 0.6177 (rounded to 4 decimal places)
horizontal speed ≈ 392.54 m/s
Finally, let's calculate the horizontal distance traveled by the package:
horizontal distance = horizontal speed x time
Plugging in the values:
horizontal distance = 392.54 m/s x 3.87 s
Now let's calculate the horizontal distance:
horizontal distance ≈ 1516.4648 m
Therefore, the package travels approximately 1516.4648 meters horizontally.