An airplane with a speed of 53.8 m/s is climbing upward at an angle of 42° with respect to the horizontal. When the plane's altitude is 840 m, the pilot releases a package.
(a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.
(b) Relative to the ground determine the angle of the velocity vector of the package just before impact.
Vo = 53.8m/s[42o]
Xo = 53.8*Cos42 = 40. m/s
Yo = 53.8*Sin42 = 36.00 m/s.
a. Yo*t + 0.5g*t^2 = 840 m.
-36*t + 4.9t^2 = 840
4.9t^2 - 36t - 840 = 0
Use Quadratic formula and get:
t = 17.3 s. = Fall time.
Dx = Xo*t = 40 * 17.3 = 692 m.
b. Y^2 = Yo^2 + 2g*h = 36^2 + 19.6*840 =
17,760
Y = 133.3 m/s.
Tan A = Y/Xo = 133.3/40 = 3.33167
A = 73.3o
To calculate the distance along the ground where the package hits the earth, we need to find the time it takes for the package to fall.
We can use the vertical motion equation:
y = yo + vot + 1/2at^2
Where:
- y = altitude (840 m)
- yo = initial altitude (0 m)
- vo = initial vertical velocity (v * sinθ)
- a = acceleration due to gravity (-9.8 m/s^2)
- t = time
Since the initial vertical velocity is given by v * sinθ, we can substitute the values:
840 = 0 + (53.8 * sin(42°)) * t + 1/2 * (-9.8) * t^2
Simplifying the equation, we get:
840 = 22.97t - 4.9t^2
To solve for t, we rearrange the equation and set it equal to zero:
4.9t^2 - 22.97t + 840 = 0
Using the quadratic formula, we can find that t ≈ 17.83 seconds.
Now, to find the horizontal distance traveled by the package, we can use the horizontal motion equation:
x = xo + vot + 1/2at^2
Since there is no horizontal acceleration, the equation becomes:
x = xo + vot
The horizontal velocity (vx) is given by v * cosθ. So, substituting the values:
x = 0 + (53.8 * cos(42°)) * 17.83
Calculating it, we find that the distance along the ground to where the package hits the earth is approximately 708 meters.
To determine the angle of the velocity vector of the package just before impact relative to the ground, we can use trigonometry. The horizontal velocity (vx) is given by v * cosθ, and the vertical velocity (vy) is given by v * sinθ.
Let's calculate the horizontal and vertical velocities first:
vx = 53.8 * cos(42°) ≈ 39.17 m/s
vy = 53.8 * sin(42°) ≈ 35.30 m/s
The angle of the velocity vector can be found using the arctan function:
θ' = arctan(vy / vx)
Substituting the values, we have:
θ' = arctan(35.30 / 39.17)
Calculating it, we find that the angle of the velocity vector just before impact is approximately 42.87° relative to the ground.