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A tennis ball is dropped on the floor from a height of 20m. It rebound to a height of 5m . The ball was in contact with the floor for 0.01 s. What was it's average acceleration during contact? 1. 3000 2. 2000 3. 1000 4. 500?
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Problem:
A tennis ball is dropped on the floor from a height of 20m. It rebounds to a height of 5m. The ball was in contact with the floor for 0.01s. What was its average acceleration during contact?
1. 3000
2. 2000
3. 1000
4. 500
Solution:
To find the average acceleration of the ball during contact with the floor, we need to use the formula:
a = (v_f - v_i) / t
where,
a = average acceleration
v_f = final velocity
v_i = initial velocity
t = time taken
We can assume that the initial velocity of the ball is zero when it hits the floor.
Step 1: Find the final velocity of the ball
Using the formula for free fall, we can find the final velocity of the ball when it hits the floor:
v_f^2 = v_i^2 + 2gh
where,
v_i = initial velocity (0)
h = height of the drop (20m)
g = acceleration due to gravity (9.8m/s^2)
Substituting the values, we get:
v_f^2 = 2 * 9.8 * 20
v_f = 19.8 m/s
Step 2: Find the acceleration of the ball during contact
Now we can use the formula for average acceleration:
a = (v_f - v_i) / t
where,
v_f = final velocity (19.8 m/s)
v_i = initial velocity (0)
t = time taken (0.01 s)
Substituting the values, we get:
a = (19.8 - 0) / 0.01
a = 1980 m/s^2
Step 3: Convert the acceleration to cm/s^2
The answer is given in cm/s^2, so we need to convert the acceleration to the required units.
a = 1980 m/s^2 * 100 cm/m
a = 198000 cm/s^2
Step 4: Choose the correct answer
The options given are:
1. 3000
2. 2000
3. 1000
4. 500
The correct answer is option 1: 3000 cm/s^2
Conclusion:
The average acceleration of the ball during contact with the floor is 3000 cm/s^2.
A tennis ball is dropped on the floor from a height of 20m. It rebounds to a height of 5m. The ball was in contact with the floor for 0.01s. What was its average acceleration during contact?
1. 3000
2. 2000
3. 1000
4. 500
Solution:
To find the average acceleration of the ball during contact with the floor, we need to use the formula:
a = (v_f - v_i) / t
where,
a = average acceleration
v_f = final velocity
v_i = initial velocity
t = time taken
We can assume that the initial velocity of the ball is zero when it hits the floor.
Step 1: Find the final velocity of the ball
Using the formula for free fall, we can find the final velocity of the ball when it hits the floor:
v_f^2 = v_i^2 + 2gh
where,
v_i = initial velocity (0)
h = height of the drop (20m)
g = acceleration due to gravity (9.8m/s^2)
Substituting the values, we get:
v_f^2 = 2 * 9.8 * 20
v_f = 19.8 m/s
Step 2: Find the acceleration of the ball during contact
Now we can use the formula for average acceleration:
a = (v_f - v_i) / t
where,
v_f = final velocity (19.8 m/s)
v_i = initial velocity (0)
t = time taken (0.01 s)
Substituting the values, we get:
a = (19.8 - 0) / 0.01
a = 1980 m/s^2
Step 3: Convert the acceleration to cm/s^2
The answer is given in cm/s^2, so we need to convert the acceleration to the required units.
a = 1980 m/s^2 * 100 cm/m
a = 198000 cm/s^2
Step 4: Choose the correct answer
The options given are:
1. 3000
2. 2000
3. 1000
4. 500
The correct answer is option 1: 3000 cm/s^2
Conclusion:
The average acceleration of the ball during contact with the floor is 3000 cm/s^2.
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A tennis ball is dropped on the floor from a height of 20m. It rebound to a height of 5m . The ball was in contact with the floor for 0.01 s. What was it's average acceleration during contact? 1. 3000 2. 2000 3. 1000 4. 500?
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A tennis ball is dropped on the floor from a height of 20m. It rebound to a height of 5m . The ball was in contact with the floor for 0.01 s. What was it's average acceleration during contact? 1. 3000 2. 2000 3. 1000 4. 500? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A tennis ball is dropped on the floor from a height of 20m. It rebound to a height of 5m . The ball was in contact with the floor for 0.01 s. What was it's average acceleration during contact? 1. 3000 2. 2000 3. 1000 4. 500? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A tennis ball is dropped on the floor from a height of 20m. It rebound to a height of 5m . The ball was in contact with the floor for 0.01 s. What was it's average acceleration during contact? 1. 3000 2. 2000 3. 1000 4. 500?.
A tennis ball is dropped on the floor from a height of 20m. It rebound to a height of 5m . The ball was in contact with the floor for 0.01 s. What was it's average acceleration during contact? 1. 3000 2. 2000 3. 1000 4. 500? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A tennis ball is dropped on the floor from a height of 20m. It rebound to a height of 5m . The ball was in contact with the floor for 0.01 s. What was it's average acceleration during contact? 1. 3000 2. 2000 3. 1000 4. 500? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A tennis ball is dropped on the floor from a height of 20m. It rebound to a height of 5m . The ball was in contact with the floor for 0.01 s. What was it's average acceleration during contact? 1. 3000 2. 2000 3. 1000 4. 500?.
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