We are given that a firework is launched vertically with an initial velocity of 25 m/s. At the maximum height, it explodes into two pieces of equal mass. One piece falls back to the ground with zero initial velocity immediately after the explosion. We are tasked with finding how much time it takes for the other piece to hit the ground after the explosion.

Step-by-step solution:

1. Determine the maximum height (H) reached by the firework before the explosion.

   The velocity at the maximum height will be zero, so we can use the kinematic equation to find the maximum height:

   $v^2 = u^2 + 2a s$

   Where:

   • $v = 0$ m/s (velocity at maximum height),
   • $u = 25$ m/s (initial velocity),
   • $a = -9.8$ m/s² (acceleration due to gravity, acting downwards),
   • $s = H$ (maximum height).

   Substituting the values into the equation: $0 = (25)^2 + 2(-9.8)H$ Solving for $H$: $625 = 19.6H$ $H = \frac{625}{19.6} = 31.8878 \, \text{m}$ So, the maximum height is approximately 31.89 meters.

2. Consider the explosion:

   The firework explodes at the maximum height (31.89 m). Since one piece falls back with zero initial velocity, it will take time to fall back to the ground, which we can calculate using the kinematic equation:

   $s = ut + \frac{1}{2} a t^2$

   For the piece that falls with zero initial velocity:

   • $u = 0$ m/s (initial velocity),
   • $a = 9.8$ m/s² (acceleration due to gravity),
   • $s = 31.89$ m (distance fallen).

   The equation becomes: $31.89 = 0 \cdot t + \frac{1}{2} (9.8) t^2$ Simplifying: $31.89 = 4.9 t^2$ Solving for $t$: $t^2 = \frac{31.89}{4.9} = 6.5$ $t = \sqrt{6.5} \approx 2.55 \, \text{seconds}$ So, it takes about 2.55 seconds for this piece to hit the ground.

3. For the other piece: After the explosion, the other piece will move upwards initially, since the explosion imparted an upward velocity. The total velocity just after the explosion for the second piece will be equal in magnitude but opposite in direction (since they have equal mass, momentum is conserved). So, the second piece will have an initial velocity of 25 m/s upwards.

   We now need to determine how long it takes for the second piece to hit the ground. The time to reach the highest point of the second piece is the same as in the first case, since it has the same initial velocity.

   The time to reach the highest point is: $t_{\text{up}} = \frac{u}{g} = \frac{25}{9.8} \approx 2.55 \, \text{seconds}$

   After reaching the maximum height, the piece will fall back down, and the time taken to fall is the same as the time it takes to reach the highest point.

   Therefore, the total time for the second piece to hit the ground is: $t_{\text{total}} = 2.55 \, \text{seconds (up)} + 2.55 \, \text{seconds (down)} = 5.10 \, \text{seconds}$

Final Answer:

The second piece takes 5.10 seconds to hit the ground after the explosion.

Would you like more details on any part of this solution? Here are some related questions:

1. How does the explosion affect the motion of the two pieces?
2. What would happen if the firework were launched with a different initial velocity?
3. How does conservation of momentum apply to this problem?
4. How would the result change if the firework were launched at an angle?
5. What role does gravity play in the motion of the pieces?

Tip: When solving problems involving explosions, always remember that momentum is conserved in the absence of external forces.