The problem describes a scenario where two fireworks, A and B, are launched at different times and speeds, and we need to determine when both fireworks will explode at the same height in the air. Here’s a structured approach to solving it:

1. Define Variables and Equations:

   • Let $t$ represent the time after Firework B's launch when both fireworks explode.
   • Since Firework B launches 0.25 seconds before Firework A, Firework A will have been in the air for $t - 0.25$ seconds when both explode.

2. Assume the Speeds:

   • The question references speeds "shown below," which isn't visible in the image provided. Generally, if we let $v_A$ be the speed of Firework A and $v_B$ the speed of Firework B, we can set up equations based on these values.

3. Set up the Height Equations:

   • The height $H$ reached by each firework can be given as $H = v \cdot t$, where $t$ is the time in the air, and $v$ is the speed.
   • Therefore, we have:
     • Height of Firework A = $v_A \cdot (t - 0.25)$
     • Height of Firework B = $v_B \cdot t$

4. Equate the Heights:

   • Since both explode at the same height, we set the heights equal to each other: $v_A \cdot (t - 0.25) = v_B \cdot t$

5. Solve for $t$:

   • Rearranging this equation will yield a solution for $t$, but we need the values of $v_A$ and $v_B$ to proceed.

Could you provide the specific speeds of Fireworks A and B? With these values, I can solve the equation for the exact time $t$.

Would you like more details on the process or have any questions? Here are some related questions that might interest you:

1. How do differences in launch speed affect synchronized timing of explosions?
2. What equations describe projectile motion in such scenarios?
3. How would a delay greater than 0.25 seconds affect the outcome?
4. What impact does air resistance have on such calculations?
5. Could this scenario be solved with calculus if acceleration were involved?

Tip: When solving these kinds of problems, always define all known variables and equations clearly before starting the solution.