A care package is released from a plane moving with a horizontal speed of 123 m/s. The plane flies at a constant height of 1013 m above the ground.
a.) Find the speed of the care package just before it hits the ground.
b.) Find the magnitude of the angle with respect to the x-direction of the care package's velocity just before it hits the ground.
I can't figure out these two, and any help would be great. Mainly, though, I am not sure how to find Vo (initial velocity) in the Y direction, or the Vf. Thanks!
a. Yf^2 = Yo^2 + 2g*h.
Yo = 0.
g = 9.8 m/s^2.
h = 1013 m.
Yf = ?.
Vf = Sqrt(Xo^2 + Yf^2), Xo = 123 m/s.
b. Tan A = Yf/Xo.
To find the speed of the care package just before it hits the ground, you need to determine the time it takes for the package to fall from the height of 1013 m.
Step 1: Determine the time of flight (t).
The formula to calculate the time of flight is:
t = √(2h/g)
where h is the height and g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
Substituting the given values:
t = √(2 * 1013 / 9.8)
t ≈ √(2066 / 9.8)
t ≈ √210.82
t ≈ 14.52 seconds
Step 2: Determine the horizontal distance traveled (x).
The horizontal distance traveled is equal to the horizontal speed of the plane multiplied by the time of flight.
x = 123 m/s * 14.52 s
x ≈ 1787.96 meters
Step 3: Determine the vertical component of velocity (Vy) just before hitting the ground.
The vertical component of velocity can be calculated as:
Vy = g * t
where g is the acceleration due to gravity.
Substituting the given values:
Vy ≈ 9.8 m/s^2 * 14.52 s
Vy ≈ 142.30 m/s
Step 4: Find the speed of the care package just before it hits the ground (V).
The speed can be calculated using the Pythagorean theorem:
V = √(Vx^2 + Vy^2)
where Vx is the horizontal component of velocity.
Substituting the given values:
V = √(123 m/s)^2 + (142.30 m/s)^2
V ≈ √15129 + 20203.29
V ≈ √35332.29
V ≈ 187.9 m/s
Therefore, the speed of the care package just before it hits the ground is approximately 187.9 m/s.
Now, let's move on to find the magnitude of the angle with respect to the x-direction of the care package's velocity just before hitting the ground.
Step 1: Determine the angle (θ) using trigonometry.
The angle can be calculated as:
θ = tan^(-1)(Vy / Vx)
Substituting the given values:
θ = tan^(-1)(142.30 m/s / 123 m/s)
θ ≈ tan^(-1)(1.1594)
θ ≈ 48.01 degrees
Therefore, the magnitude of the angle with respect to the x-direction of the care package's velocity just before it hits the ground is approximately 48.01 degrees.
To find the speed of the care package just before it hits the ground, we need to determine its final vertical velocity. We can use the equation of motion in the vertical direction:
Final vertical velocity (Vf) = Initial vertical velocity (Vo) + (Acceleration due to gravity × Time)
In this case, the initial vertical velocity, Vo, is zero because the care package is released from rest. The acceleration due to gravity is approximately 9.8 m/s², and the time it takes for the care package to hit the ground can be found using the equation:
Time = Distance / Horizontal velocity, where the distance is the height of the plane above the ground, which is given as 1013 m, and the horizontal velocity is given as 123 m/s.
Let's calculate the time it takes for the care package to hit the ground:
Time = 1013 m / 123 m/s
Time ≈ 8.22 s
Now, we can substitute the values into the equation for Vf:
Vf = 0 + (9.8 m/s² × 8.22 s)
Vf ≈ 80.6 m/s
Therefore, the speed of the care package just before it hits the ground is approximately 80.6 m/s.
To find the magnitude of the angle with respect to the x-direction of the care package's velocity just before it hits the ground, we can use trigonometry. The horizontal velocity remains constant at 123 m/s, and the vertical velocity just before hitting the ground is -80.6 m/s (negative due to downward direction). Therefore, we can calculate the angle θ as follows:
θ = arctan(V_vertical / V_horizontal)
θ = arctan(-80.6 m/s / 123 m/s)
θ ≈ -29.2°
The magnitude of the angle with respect to the x-direction of the care package's velocity just before it hits the ground is approximately 29.2°. Note that the negative sign indicates that the angle is measured below the x-axis.