a ball is dropped from height 10 m. it rebounds to a height of 5m.the ball was in contact with the floor for 0.01s what was its acceleration during contact?
u = root ( 2gh)
u = root(2 * 9.8 * 10)
u = root(196
u = 14 m / s
v = root(2gh)
v = root(2*9.8 * 5)
v = root(98)
v = root 98
v = 7 root2
14 + 7 root2
Acceleration = (14+7root2)/ 0.01
= 7(2 + root2) / 0.01
= 700(2 + root2)
700(2 + 1.4142)
= 700(3.4142)
= 341.42 *7
= 2389.94 m/sec^2
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b
c
bc
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59m/s^2
To calculate the acceleration during the contact of the ball with the floor, we can use the formula:
acceleration = (final velocity - initial velocity) / time
In this case, we need to find the change in velocity during the contact. Since the ball was dropped and rebounds, its initial velocity is 0 m/s (when it starts from rest) and the final velocity is determined by the rebound height.
The formula for calculating the final velocity during an elastic collision is given by:
final velocity = √(2gh)
Where:
g = acceleration due to gravity (approximately 9.8 m/s²)
h = rebound height (in this case, 5 m)
So, the final velocity can be calculated as follows:
final velocity = √(2 * 9.8 * 5) m/s
final velocity = √(98) m/s
final velocity ≈ 9.90 m/s
Now, we can substitute the values into the acceleration formula:
acceleration = (final velocity - initial velocity) / time
acceleration = (9.90 - 0) / 0.01 m/s²
acceleration = 9.90 / 0.01 m/s²
acceleration = 990 m/s²
Therefore, the acceleration during the contact of the ball with the floor is approximately 990 m/s².