An airplane with a speed of 13 m/s is climbing upward at an angle of 48° counterclockwise from the positive x axis. When the plane's altitude is 720 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) Determine the angle of the velocity vector of the package just before impact.
2° clockwise from the positive x axis
To solve this problem, we need to consider the motion of the airplane and the package separately.
For part (a), we need to find the horizontal distance traveled by the package when it hits the ground. To do this, we'll first find the time it takes for the package to reach the ground. We'll then use this time to calculate the horizontal distance.
Step 1: Find the time of flight
Given:
Initial vertical velocity of the package (v₀y) = 0 m/s (since the package is released when the airplane is at its maximum altitude)
Acceleration due to gravity (g) = 9.8 m/s²
Using the kinematic equation for vertical motion:
v = u + at
0 = v₀y + (-g)t (downward direction is taken as negative)
t = v₀y / g
t = 0 / 9.8
t = 0 seconds (tells us that the package is already at the ground)
Step 2: Find the horizontal distance
Since time is zero, the package doesn't travel horizontally from the point of release. Therefore, the distance along the ground is also zero.
So, the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth is 0 meters.
For part (b), we need to find the angle of the velocity vector of the package just before impact. Since the package is falling vertically, its velocity vector will have a downward direction.
Given:
Angle of the airplane's velocity vector (θ) = 48° counterclockwise from the positive x-axis
Angle of the velocity vector of the package just before impact = 2° clockwise from the positive x-axis
By subtracting the two angles, we can find the angle of the velocity vector of the package just before impact:
θ_package = θ_airplane - 180° + α (where α is the angle of the velocity vector of the package)
θ_package = 48° - 180° + (-2°)
θ_package = -134°
So, the angle of the velocity vector of the package just before impact is -134° clockwise from the positive x-axis.