We can solve this problem by applying the equations of motion. The ball is thrown upward with an initial velocity and will eventually fall to the ground due to gravity. The vertical motion of the ball can be modeled using the following kinematic equation:

$y(t) = y_0 + v_0t - \frac{1}{2}gt^2$

Where:

• $y(t)$ is the position of the ball at time $t$,
• $y_0 = 25 \, \text{m}$ is the initial height of the ball (the height of the building),
• $v_0 = 15 \, \text{m/s}$ is the initial velocity (upward),
• $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity (downward),
• $t$ is the time.

We are trying to find the time $t$ when the ball reaches the ground. When it hits the ground, its height is zero ($y(t) = 0$).

Thus, the equation becomes:

$0 = 25 + 15t - \frac{1}{2}(9.8)t^2$

This is a quadratic equation of the form:

$0 = -4.9t^2 + 15t + 25$

We can solve this quadratic equation for $t$ using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where $a = -4.9$, $b = 15$, and $c = 25$.

Let's calculate the time $t$.The two solutions for time $t$ are approximately $-1.20 \, \text{seconds}$ and $4.26 \, \text{seconds}$. Since time cannot be negative, we discard the negative solution.

Thus, the ball will take approximately 4.26 seconds to reach the ground.

Would you like further details on the solution or have any other questions?

Here are 5 related questions to expand your understanding:

1. What is the maximum height the ball reaches before falling back down?
2. How long does it take for the ball to reach its maximum height?
3. What is the velocity of the ball just before it hits the ground?
4. What would change if the initial speed were doubled?
5. How would the total time change if the building were taller?

Tip: Always check the direction of motion (upward or downward) when applying kinematic equations!