A coconut falls from a 12m tree. What be its velocity halfway down?

To find the velocity of the coconut halfway down the tree, we can use the equation of motion:

v^2 = u^2 + 2as

Where:
v = final velocity
u = initial velocity (0 m/s as the coconut starts from rest)
a = acceleration due to gravity (9.81 m/s^2)
s = distance fallen (6m, halfway down the tree)

Plugging in the values, we get:

v^2 = 0 + 2(9.81)(6)
v^2 = 117.72
v = √117.72
v ≈ 10.85 m/s

Therefore, the velocity of the coconut halfway down the tree is approximately 10.85 m/s.

Step 1: Determine the acceleration due to gravity (g)

```
g = 9.8 m/s²
```

Step 2: Calculate the perpendicular distance (h) traveled by the coconut halfway down

```
h = 12 m / 2 = 6 m
```

Step 3: Use the equation of motion for an object falling freely under gravity

```
v² = u² + 2gh
```

where:

- v is the velocity of the coconut halfway down
- u is the initial velocity (which is zero, as the coconut starts from rest)
- g is the acceleration due to gravity
- h is the perpendicular distance traveled

Step 4: Substitute the values into the equation

```
v² = 0² + 2 - 9.8 m/s² - 6 m
v² = 117.6 m²/s²
```

Step 5: Solve for v

```
v = √(117.6 m²/s²)
v = 10.8 m/s
```

Therefore, the velocity of the coconut halfway down is 10.8 m/s.