An airplane with a speed of 88.2 m/s is climbing upward at an angle of 55.6 ° with respect to the horizontal. When the plane's altitude is 523 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.

Question

An airplane with a speed of 88.2 m/s is climbing upward at an angle of 55.6 ° with respect to the horizontal. When the plane's altitude is 523 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.

Answer 1

Vi = +88.2 sin 55.6 = 72.8

u = 88.2 cos 55.6 until it hits the ground = 49.8

v = Vi - 9.81 t
h = Hi + Vi t - 4.9 t^2

so when is h = 0 ?
0 = 523 + 72.8 t - 4.9 t^2
solve quadratic
t = 20.2 seconds (or -5.3 seconds but that was when it might have been thrown from the ground :)
https://www.mathsisfun.com/quadratic-equation-solver.html
x = u t = 49.8*20.2 = 1007 meters or about a kilometer
v = Vi - 9.81 (20.2) = -125 m/s
u is still 88.2
so tan angle down from level = 125/88.2
angle down from level = 54.8

Answer 2

x = u t = 49.8*20.2 = 1007 meters or about a kilometer

v = Vi - 9.81 (20.2) = -125 m/s

u is still 49.8 NOT 88.2 !!!!!!!

so tan angle down from level = 125/49.8
angle down from level = 68.3 deg

Answer 3

To solve this problem, we can break it down into two parts: (a) Calculate the distance along the ground where the package hits the earth, and (b) Determine the angle of the velocity vector of the package just before impact.

(a) 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 reach the ground. We'll use the vertical motion equation:

h = v₀t + (1/2)gt²

Where:
h = altitude = 523 m
v₀ = initial vertical velocity = 88.2 m/s * sin(55.6°)
t = time
g = acceleration due to gravity = 9.8 m/s²

We need to solve this equation for t, so let's rearrange it:

523 = (88.2 * sin(55.6°))t - (1/2)(9.8)t²

Now we have a quadratic equation. We can simplify it and solve for t using either the quadratic formula or factoring.

Once we find the value of t, we can calculate the horizontal distance using the horizontal velocity of the plane:

d = v₀x * t

Where:
d = horizontal distance
v₀x = initial horizontal velocity = 88.2 m/s * cos(55.6°)
t = time calculated earlier

(b) To determine the angle of the velocity vector of the package just before impact, we can use the following trigonometric identity:

tan(θ) = v₀y / v₀x

Where:
θ = angle of the velocity vector
v₀y = initial vertical velocity = 88.2 m/s * sin(55.6°)
v₀x = initial horizontal velocity = 88.2 m/s * cos(55.6°)

We can calculate θ using the inverse tangent (tan⁻¹) function:

θ = tan⁻¹(v₀y / v₀x)

Now we have the answer to both parts of the question.