A plane dives at 34∘ to the horizontal and releases a package at an altitude of 490 m. If the load is in the air for 6.4 s, find:
a) the speed of the plane when it released the package;
b) the horizontal distance traveled by the package after it is released.
If the plane's speed is v, then
the package's initial downward speed is v sin(-34°)
So, if it takes 6.4s to land, then
490 - v sin34° * 6.4 - 4.9*6.4^2 = 0
v = 80.84 m/s
The horizontal speed is 80.84 cos34° = 67 m/s
distance = speed * time = 67 * 6.4 = 428.8 m
To solve this problem, we can break it down into two parts:
a) To find the speed of the plane when it released the package, we need to use the horizontal and vertical components of its motion. We can use the following equations:
Horizontal component of velocity (Vx) = V0x, where V0x is the initial horizontal velocity of the plane
Vertical component of velocity (Vy) = V0y - gt, where V0y is the initial vertical velocity of the plane, g is the acceleration due to gravity (9.8 m/s^2), and t is the time the package is in the air
Since the plane is diving at an angle of 34∘ to the horizontal, we can calculate the initial horizontal and vertical velocities:
V0x = Vp cosθ = Vp cos34∘
V0y = Vp sinθ = Vp sin34∘
where Vp is the speed of the plane.
Now, we can find Vp by using the equation for vertical displacement:
Vertical displacement (Δy) = V0y * t - (1/2) * g * t^2
Δy = 490 m
t = 6.4 s
g = 9.8 m/s^2
Plugging in these values, we can solve for Vp:
490 = (Vp sin34∘) * 6.4 - (1/2) * 9.8 * (6.4^2)
b) To find the horizontal distance traveled by the package after it is released, we need to use the horizontal velocity of the plane and the time the package is in the air. We can use the equation:
Horizontal displacement (Δx) = V0x * t
Now, we can solve for Δx by plugging in the values we have:
Δx = (Vp cos34∘) * 6.4
By solving these equations, we can find both the speed of the plane when it released the package (a) and the horizontal distance traveled by the package (b).
To solve this problem, we can use the equations of motion to analyze the motion of the package. Let's break it down step by step:
a) First, let's determine the vertical motion of the package. We can use the equation:
y = y0 + v0y*t - 0.5*g*t^2
where:
y is the vertical displacement of the package (490 m),
y0 is the initial vertical position (0 m),
v0y is the initial vertical velocity (unknown),
t is the time (6.4 s),
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the package is released vertically, the initial vertical velocity (v0y) is 0. Plugging in the given values, we have:
490 m = 0 m + 0 m/s * 6.4 s - 0.5 * 9.8 m/s^2 * (6.4 s)^2
Simplifying this equation gives us:
490 m = -0.5 * 9.8 m/s^2 * (6.4 s)^2
Now solve the equation for (6.4 s)^2:
(6.4 s)^2 = (490 m) / (-0.5 * 9.8 m/s^2)
(6.4 s)^2 = 1020 s^2
Taking the square root of both sides, we find:
6.4 s = √1020 s ≈ 31.95 s
Therefore, the time taken to reach 490 m in the vertical direction is approximately 31.95 seconds.
b) Now, let's determine the horizontal component of the plane's velocity when it released the package. We can use the equation:
v = v0x
where:
v is the horizontal component of the velocity (unknown),
v0x is the initial horizontal component of the velocity (unknown).
We know that the angle of dive of the plane is 34 degrees to the horizontal. This means that the vertical velocity (v0y) is related to the horizontal velocity (v0x) through the following equation:
v0y = v0x * tan(34°)
Since the initial vertical velocity (v0y) is 0 (because the package was released vertically), we can solve for v0x:
v0x * tan(34°) = 0
Simplifying the equation gives us:
v0x = 0 m/s
Therefore, the initial horizontal velocity (v0x) of the plane is 0 m/s.
c) Lastly, let's calculate the horizontal distance traveled by the package after it is released. We can use the equation:
x = v0x * t
where:
x is the horizontal distance (unknown),
v0x is the initial horizontal velocity (0 m/s),
t is the time (6.4 s).
Plugging in the values, we have:
x = 0 m/s * 6.4 s = 0 m
Therefore, the horizontal distance traveled by the package after it is released is 0 meters.
In summary:
a) The speed of the plane when it released the package is 0 m/s.
b) The horizontal distance traveled by the package after it is released is 0 meters.