A ball is dropped from a height h. After 1 s, another ball is dropped from a point which is 25 m vertically below the point from where the first ball was dropped. If both the balls reach the ground at the same time, then distance travelled by 1st ball in last second of its motion (g=10 m/s2)

The problem involves the equations of motion under constant acceleration due to gravity.

1. Let's first find the total time it takes for the first ball to hit the ground. We know that the distance a falling object travels is given by the equation d = 0.5gt^2, where g is the acceleration due to gravity and t is the time.

   So, h = 0.5 * 10 * t^2 => t^2 = h/5 => t = sqrt(h/5)

2. We know that the second ball is dropped 1 second after the first, and they both hit the ground at the same time. This means the second ball is in the air for 1 second less than the first ball.

   So, the time for the second ball, t2 = t - 1

3. We also know that the second ball is dropped from a height 25m less than the first ball. So, the height from which the second ball is dropped, h2 = h - 25

4. We can now set up the equation for the second ball's fall and solve for h.

   h2 = 0.5 * 10 * t2^2 => h - 25 = 5 * (t - 1)^2

5. We can solve this equation for h, and then substitute this value back into the equation for t to find the total time the first ball is in the air.

6. The distance travelled by the first ball in the last second of its motion can be found using the equation d = gt, where t is 1 second.

   So, d = 10 * 1 = 10m

So, the first ball travels 10m in the last second of its motion.

The problem involves the equations of motion.

Let's denote:

• the time it takes for the first ball to hit the ground as T
• the height from which the first ball is dropped as h
• the height from which the second ball is dropped as h - 25 m
• the acceleration due to gravity as g = 10 m/s^2

From the problem, we know that the second ball is dropped 1 second after the first ball, so it hits the ground in T - 1 seconds.

We can use the equation of motion: h = 0.5 * g * T^2 for the first ball and h - 25 = 0.5 * g * (T - 1)^2 for the second ball.

Solving these two equations, we get T = 6 seconds.

The distance travelled by the first ball in the last second of its motion can be calculated using the equation of motion: s = u + 0.5 * g * t^2, where u is the initial velocity, g is the acceleration due to gravity, t is the time and s is the distance travelled.

In the last second of its motion, the initial velocity of the first ball is u = g * (T - 1) = 10 * 5 = 50 m/s. The time is t = 1 second.

Substituting these values into the equation, we get s = 50 + 0.5 * 10 * 1^2 = 55 m.

So, the first ball travels 55 m in the last second of its motion.