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We are given two scenarios with two balls: one is thrown upward from the ground, and the other is dropped from a height of 15 m. We need to determine when they will be at the same height.

Let's define the key parameters:

1. **Ball 1 (thrown upward)**:
   - Initial velocity, \( u_1 = 25 \, \text{m/s} \)
   - Initial height, \( h_1 = 0 \, \text{m} \)
   - Acceleration, \( a_1 = -9.8 \, \text{m/s}^2 \) (due to gravity acting downward)

   The height at time \( t \) for Ball 1 can be described by the kinematic equation:
   \[
   y_1(t) = u_1 t + \frac{1}{2} a_1 t^2
   \]
   Substituting values:
   \[
   y_1(t) = 25t - 4.9t^2
   \]

2. **Ball 2 (dropped from height)**:
   - Initial velocity, \( u_2 = 0 \, \text{m/s} \)
   - Initial height, \( h_2 = 15 \, \text{m} \)
   - Acceleration, \( a_2 = -9.8 \, \text{m/s}^2 \) (due to gravity)

   The height at time \( t \) for Ball 2 can be described by:
   \[
   y_2(t) = h_2 + u_2 t + \frac{1}{2} a_2 t^2
   \]
   Substituting values:
   \[
   y_2(t) = 15 - 4.9t^2
   \]

### Solution:

Now, we set \( y_1(t) = y_2(t) \) to find the time when the heights are equal:

\[
25t - 4.9t^2 = 15 - 4.9t^2
\]

Simplifying, the \( -4.9t^2 \) terms cancel out, leaving:
\[
25t = 15
\]

Solving for \( t \):
\[
t = \frac{15}{25} = 0.6 \, \text{seconds}
\]

Thus, the two balls will be at the same height after **0.60 seconds**.

Would you like more details on this problem or have any questions?

### Related questions:
1. How do you derive the kinematic equation for constant acceleration?
2. What would the situation look like if air resistance was considered?
3. How high will the first ball go before starting to fall back?
4. What is the velocity of the first ball at the moment they are at the same height?
5. How can we calculate the height at which the two balls meet?

### Tip:
When solving for time in kinematic equations, notice how simplifying steps (like canceling terms) can often make complex-looking problems easier to handle.

A ball is thrown upward from the ground with an initial speed of 25 m/s; at the same instant, a ball is dropped from rest from a building 15 m high. After how long will the balls be at the same height?

Explore how to calculate the time when two balls—one thrown upward with an initial speed of 25 m/s and another dropped from a 15-meter height—reach the same height. Learn about kinematic equations and their application to solve this problem, ideal for students in Grades 11-12.

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