Mr. X drops A ball from a height of 1 metre above the ground. Mr. Y starts a camera which takes picture of the ball every 250 ms, with the first click occuring at the moment Mr. X lets go of the ball. Let the ball's height about the ground at the n-th click, in metres, be h_n. Compute the value of [(h_3+h_4+h_5)/h_2]

Assumptions: If the ball makes a collision with the ground, its totally elastic.

To find the value of [(h_3+h_4+h_5)/h_2], we need to determine the heights of the ball at different time intervals.

Starting from the initial drop, we'll break down the problem step by step:

1. The ball is dropped from a height of 1 meter, so at the first click (t = 0), the height of the ball is 1 meter (h_1 = 1).

2. The camera takes pictures every 250 milliseconds, so at the second click (t = 250 ms), the ball has been falling for 250 ms. To calculate the height at this time, we need to determine how much the ball has fallen. We can use the equation of motion for free fall:
h = h_0 - (1/2)gt^2

Here, h_0 is the initial height (1 meter), g is the acceleration due to gravity (9.8 m/s^2), and t is the time in seconds (0.25 s).
Plugging in the values, we get:
h_2 = 1 - (1/2)(9.8)(0.25)^2 = 1 - 0.30625 = 0.69375 meters.

3. At the third click (t = 500 ms), the ball has been falling for 500 ms. Again, using the same equation of motion:
h_3 = 1 - (1/2)(9.8)(0.5)^2 = 1 - 0.5 = 0.5 meters.

4. At the fourth click (t = 750 ms), the ball has been falling for 750 ms:
h_4 = 1 - (1/2)(9.8)(0.75)^2 = 1 - 0.275625 = 0.724375 meters.

5. At the fifth click (t = 1000 ms), the ball has been falling for 1000 ms:
h_5 = 1 - (1/2)(9.8)(1)^2 = 1 - 4.9 = -3.9 meters.

Because the ball has collided with the ground at the fifth click, the height becomes negative (-3.9 meters).

Now, we can calculate the value of [(h_3+h_4+h_5)/h_2]:
[(h_3+h_4+h_5)/h_2] = [(0.5 + 0.724375 - 3.9)/0.69375] = [-2.675625/0.69375] = -3.85714.