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A ball is thrown up with an initial velocity of 20 m/s and after some time it returns. What is the maximum height reached? Take g = 10 m/s2.
- a)80m
- b)20m
- c)70m
- d)40m
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A ball is thrown up with an initial velocity of 20 m/s and after some ...
Here we will use the third equation of motion. The third equation of motion is v2 = u2 + 2as. At the maximum height, the velocity of the ball will be zero, therefore, v = 0, u = 20, and a = -g = -10. On solving, we will get s = 20m.
Most Upvoted Answer
A ball is thrown up with an initial velocity of 20 m/s and after some ...
Understanding the Problem
To find the maximum height reached by the ball, we can use the following kinematic equation:
\[
v^2 = u^2 + 2as
\]
Where:
- \( v \) = final velocity (0 m/s at the maximum height)
- \( u \) = initial velocity (20 m/s)
- \( a \) = acceleration (due to gravity, -10 m/s², since it acts downward)
- \( s \) = displacement (maximum height reached)
Applying the Values
At the maximum height, the final velocity \( v = 0 \) m/s. Plugging in the values into the equation:
\[
0 = (20)^2 + 2(-10)s
\]
This simplifies to:
\[
0 = 400 - 20s
\]
Rearranging gives:
\[
20s = 400
\]
Thus, solving for \( s \):
\[
s = \frac{400}{20} = 20 \text{ m}
\]
Conclusion
The maximum height reached by the ball is 20 meters.
Correct Answer
Therefore, the correct answer is option B (20 m).
To find the maximum height reached by the ball, we can use the following kinematic equation:
\[
v^2 = u^2 + 2as
\]
Where:
- \( v \) = final velocity (0 m/s at the maximum height)
- \( u \) = initial velocity (20 m/s)
- \( a \) = acceleration (due to gravity, -10 m/s², since it acts downward)
- \( s \) = displacement (maximum height reached)
Applying the Values
At the maximum height, the final velocity \( v = 0 \) m/s. Plugging in the values into the equation:
\[
0 = (20)^2 + 2(-10)s
\]
This simplifies to:
\[
0 = 400 - 20s
\]
Rearranging gives:
\[
20s = 400
\]
Thus, solving for \( s \):
\[
s = \frac{400}{20} = 20 \text{ m}
\]
Conclusion
The maximum height reached by the ball is 20 meters.
Correct Answer
Therefore, the correct answer is option B (20 m).
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A ball is thrown up with an initial velocity of 20 m/s and after some time it returns. What is the maximum height reached? Take g = 10 m/s2.a)80mb)20mc)70md)40mCorrect answer is option 'B'. Can you explain this answer?
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