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 groundat 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 compute the value of [(h_3+h_4+h_5)/h_2], we first need to determine the heights of the ball at each click.
Since the ball is dropped from a height of 1 meter, the height of the ball at the first click (h_1) is 1 meter.
The time interval between each click is 250 ms, which is equivalent to 0.25 seconds. Since the ball is in freefall, it accelerates due to gravity, and the distance it falls is given by the formula:
d = (1/2)gt^2
where d is the distance fallen, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time in seconds.
To find the height at the second click (h_2), we substitute t = 0.25 seconds into the formula:
h_2 = 1 - (1/2)(9.8)(0.25)^2
= 1 - (1/2)(9.8)(0.0625)
= 1 - 0.305
= 0.695 meters
To find the heights at the third, fourth, and fifth clicks, we need to calculate the distance fallen at each interval.
For the third click (h_3), the time is 0.5 seconds (2 x 0.25 seconds):
h_3 = 1 - (1/2)(9.8)(0.5)^2
= 0.51 meters
For the fourth click (h_4), the time is 0.75 seconds (3 x 0.25 seconds):
h_4 = 1 - (1/2)(9.8)(0.75)^2
= 0.117 meters
For the fifth click (h_5), the time is 1 second (4 x 0.25 seconds):
h_5 = 1 - (1/2)(9.8)(1)^2
= 0 meters (hits the ground, but bounces back up to a certain height due to the elastic collision)
Now that we have determined the heights at each click, we can calculate the desired value:
[(h_3+h_4+h_5)/h_2] = [(0.51+0.117+0)/0.695]
= 0.728
Therefore, the value of [(h_3+h_4+h_5)/h_2] is 0.728.