Dynamics, plane drops mass trajectory question

[cd80187]

Homework Statement

The pilot of an airplane, carrying a package of mail, wishes to hit recovery location A. Assume that the plane is traveling at constant speed of 200 km/hr at a constant height of 100m above the ground. What angle theta with the horizontal should the pilot's line of sight to the target make at the instant of release?

Homework Equations

I don't understand what equations to use, if you could help that would be awesome.

The Attempt at a Solution

 

[teknodude]

Just start with the equations of motion under gravity, or you could use the already derived free fall equations, but its better start from scratch and derive it. They tell you that speed is constant, so what are the acceleration components?

 

[piercas]

To solve this problem, we can use the equation for projectile motion: h = h0 + v0sin(theta)t - (1/2)gt^2, where h is the height, h0 is the initial height, v0 is the initial velocity, theta is the angle with the horizontal, t is time, and g is the acceleration due to gravity.

In this case, we know that h0 = 100m, v0 = 200 km/hr = 55.6 m/s, and g = 9.8 m/s^2. We also know that the plane will drop the package at the instant of release, so the time t will be the same for both the horizontal and vertical components of the motion.

Since we want the package to hit recovery location A, which is on the ground at a horizontal distance x from the plane's initial position, we can use the equation x = v0cos(theta)t to find the time t.

Substituting this value of t into the first equation, we can solve for theta:

100m = 100m + (55.6 m/s)sin(theta)t - (1/2)(9.8 m/s^2)t^2

0 = (55.6 m/s)sin(theta)t - (1/2)(9.8 m/s^2)t^2

Using the quadratic formula, we can solve for t:

t = [-(55.6 m/s)sin(theta) ± √((55.6 m/s)sin(theta))^2 + 4(1/2)(9.8 m/s^2)(100m)] / 2(1/2)(9.8 m/s^2)

Simplifying, we get:

t = [-(55.6 m/s)sin(theta) ± √(3088.16sin^2(theta) + 19600)] / 9.8 m/s

Since we want the time t to be positive, we can ignore the negative solution. Therefore, we have:

t = [-(55.6 m/s)sin(theta) + √(3088.16sin^2(theta) + 19600)] / 9.8 m/s

Now, we can plug this value of t into the equation x = v0cos(theta)t to solve for theta:

x = (55.6 m/s)cos(theta)[-(55.