To solve this problem, we will use the equations of motion under constant acceleration due to gravity.

Given:

• The time taken for the ball to hit the ground, $t = 2.6$ seconds.
• The acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$ (assuming the motion is happening on Earth).

1. Final Velocity of the Ball

The final velocity $v$ of an object falling freely from rest can be calculated using the equation: $v = u + gt$ where:

• $u$ is the initial velocity (which is $0 \, \text{m/s}$ since the ball was dropped),
• $g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$),
• $t$ is the time in seconds.

Substitute the values into the equation:

$v = 0 + (9.8 \, \text{m/s}^2)(2.6 \, \text{s})$ $v = 25.48 \, \text{m/s}$

So, the final velocity of the ball just before it hits the ground is 25.48 m/s.

2. Height of the Building

The height $h$ from which the ball falls can be determined using the equation: $h = ut + \frac{1}{2}gt^2$

Since the initial velocity $u = 0$, the equation simplifies to: $h = \frac{1}{2}gt^2$

Substitute the values into the equation:

$h = \frac{1}{2} (9.8 \, \text{m/s}^2)(2.6 \, \text{s})^2$ $h = \frac{1}{2} (9.8)(6.76)$ $h = 33.148 \, \text{m}$

So, the height of the building is 33.148 meters.

Summary:

• The final velocity of the ball just before it hits the ground is 25.48 m/s.
• The height of the building is 33.148 meters.

Would you like further details or have any questions?

Here are some related questions for further exploration:

1. What would be the final velocity if the ball was thrown downwards instead of being dropped?
2. How would the final velocity change if the ball was thrown upwards initially?
3. What is the potential energy of the ball at the top of the building before it falls?
4. How long would it take for the ball to fall if the building were twice as high?
5. What factors could affect the accuracy of these calculations in a real-world scenario?

Tip: Remember that in free fall, the only force acting on an object (neglecting air resistance) is gravity, which causes the object to accelerate downwards at $9.8 \, \text{m/s}^2$ on Earth.