Your friend is an environmentalist who is living in a tree for the summer. You are helping provide her with food, and you do so by throwing small packages up to her tree house. If her tree house is 21 m above the ground, what is the minimum (initial) speed you must use when throwing packages up to her?

To find the minimum initial speed required to throw the packages up to her tree house, we can use the principles of projectile motion.

The key is to determine the vertical velocity component needed for the packages to reach a height of 21 m.

In projectile motion, the vertical motion is affected by the force of gravity. The equation for calculating the final vertical velocity (vf) is given by:

vf^2 = vi^2 + 2gh

Where:
- vf is the final vertical velocity (which is 0 when the package reaches its maximum height)
- vi is the initial vertical velocity
- g is the acceleration due to gravity (9.8 m/s^2)
- h is the height reached (21 m in this case)

Since the final vertical velocity is zero at the maximum height, we can simplify the equation to:

0 = vi^2 + 2gh

Rearranging the equation, we can solve for the initial vertical velocity:

vi^2 = -2gh

vi = √(-2gh)

Substituting the given values:

vi = √(-2 * 9.8 * 21)

Calculating this expression, we get:
vi ≈ 19.34 m/s

So, the minimum initial speed you must use when throwing packages up to her tree house is approximately 19.34 m/s.

To determine the minimum initial speed required to throw the packages up to your friend's tree house, we need to consider the laws of physics, specifically projectile motion. The initial speed should be sufficient for the package to reach a height of 21 m above the ground.

The key concept here is that when the package reaches the maximum height, its vertical velocity component will be zero. We can use this point as a reference to calculate the initial speed.

Let's assume that the package is thrown vertically upward with an initial speed, v. The acceleration due to gravity, g, will act in the downward direction. Therefore, we have:

Final vertical velocity, vf = 0 m/s (at maximum height)
Initial vertical velocity, vi = ?
Acceleration due to gravity, g = 9.8 m/s² (approximate value)

Using the equations of motion for vertical motion, we have:

vf = vi - gt,

Since the package comes to rest at the maximum height:

0 = vi - g * t_maxheight,

where t_maxheight is the time taken to reach the maximum height. We can solve this equation to find t_maxheight:

t_maxheight = vi / g.

Next, we will use the kinematic equation for vertical displacement:

∆y = vi * t - 0.5 * g * t²,

where ∆y is the displacement (height) of 21 m. Substituting the known values:

21 m = (vi / g) * t_maxheight - 0.5 * g * (vi / g)²,

Simplifying the equation:

21 m = (vi² / g) - 0.5 * vi² / g,

Multiply both sides by g:

21 m * g = vi² - 0.5 * vi²,

21 m * g = 0.5 * vi²,

2 * 21 m * g = vi²,
42 m * 9.8 m/s² = vi²,

vi = √(42 m * 9.8 m/s²),

vi ≈ 27.7 m/s.

Therefore, the minimum initial speed you must use when throwing packages up to her tree house is approximately 27.7 m/s.