To find out how long it takes for the ball to reach the ground, we can use the following kinematic equation:

$y = y_0 + v_0 t + \frac{1}{2} a t^2$

Where:

• $y$ is the final position (which is 0 meters since it hits the ground).
• $y_0 = 25$ m is the initial height of the ball from the ground (the height of the building).
• $v_0 = 15$ m/s is the initial velocity (upward).
• $a = -9.8$ m/s² is the acceleration due to gravity (negative since it's acting downward).
• $t$ is the time we are trying to find.

We need to solve for $t$. Plugging in the values:

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

This simplifies to the quadratic equation:

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

Now let's 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 compute this next.The two solutions for $t$ are approximately $-1.20$ seconds and $4.26$ seconds. Since time cannot be negative, the valid solution is:

$t \approx 4.26 \, \text{seconds}$

Thus, it takes approximately 4.26 seconds for the ball to hit the ground.

Would you like more details on any step of the solution?

Here are 5 related questions:

1. How high does the ball go before starting to fall back down?
2. What is the velocity of the ball just before it hits the ground?
3. What would happen if the initial speed were doubled?
4. How would the time change if the building were taller?
5. Can we find the time to reach the maximum height?

Tip: When solving kinematic problems, always choose your sign convention carefully (positive and negative directions).