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A ball is thrown vertically upward with a speed v from a height h metre above the ground. The time taken for the ball to hit ground iscorrect answer.: v/g[1 √1 √2hg/v×v]?how to solve?
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Understanding the Problem
To determine the time taken for a ball thrown vertically upward with an initial speed \( v \) from a height \( h \) to hit the ground, we can use the equations of motion under gravity.
Key Parameters
- Initial speed: \( v \)
- Initial height: \( h \)
- Acceleration due to gravity: \( g \) (acting downward)
Equation of Motion
The vertical motion can be described by the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s \) is the displacement (which will be \(-h\) when the ball hits the ground),
- \( u \) is the initial speed (\( v \)),
- \( a \) is the acceleration (\(-g\)),
- \( t \) is the time taken.
Setting Up the Equation
When the ball is thrown upward and then falls to the ground, we can set our displacement as:
\[ -h = vt - \frac{1}{2} g t^2 \]
Rearranging gives:
\[ \frac{1}{2} g t^2 - vt - h = 0 \]
Using the Quadratic Formula
This is a quadratic equation in \( t \). We can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = \frac{1}{2} g \),
- \( b = -v \),
- \( c = -h \).
Substituting these values:
\[ t = \frac{v \pm \sqrt{v^2 + 2gh}}{g} \]
Choosing the Correct Root
We discard the negative root since time cannot be negative:
\[ t = \frac{v + \sqrt{v^2 + 2gh}}{g} \]
Thus, the time taken for the ball to hit the ground is:
\[ t = \frac{v + \sqrt{v^2 + 2gh}}{g} \]
This equation gives a clear understanding of how initial speed and height influence the time of flight until the ball reaches the ground.
To determine the time taken for a ball thrown vertically upward with an initial speed \( v \) from a height \( h \) to hit the ground, we can use the equations of motion under gravity.
Key Parameters
- Initial speed: \( v \)
- Initial height: \( h \)
- Acceleration due to gravity: \( g \) (acting downward)
Equation of Motion
The vertical motion can be described by the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s \) is the displacement (which will be \(-h\) when the ball hits the ground),
- \( u \) is the initial speed (\( v \)),
- \( a \) is the acceleration (\(-g\)),
- \( t \) is the time taken.
Setting Up the Equation
When the ball is thrown upward and then falls to the ground, we can set our displacement as:
\[ -h = vt - \frac{1}{2} g t^2 \]
Rearranging gives:
\[ \frac{1}{2} g t^2 - vt - h = 0 \]
Using the Quadratic Formula
This is a quadratic equation in \( t \). We can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = \frac{1}{2} g \),
- \( b = -v \),
- \( c = -h \).
Substituting these values:
\[ t = \frac{v \pm \sqrt{v^2 + 2gh}}{g} \]
Choosing the Correct Root
We discard the negative root since time cannot be negative:
\[ t = \frac{v + \sqrt{v^2 + 2gh}}{g} \]
Thus, the time taken for the ball to hit the ground is:
\[ t = \frac{v + \sqrt{v^2 + 2gh}}{g} \]
This equation gives a clear understanding of how initial speed and height influence the time of flight until the ball reaches the ground.
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A ball is thrown vertically upward with a speed v from a height h metre above the ground. The time taken for the ball to hit ground iscorrect answer.: v/g[1 √1 √2hg/v×v]?how to solve?
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A ball is thrown vertically upward with a speed v from a height h metre above the ground. The time taken for the ball to hit ground iscorrect answer.: v/g[1 √1 √2hg/v×v]?how to solve? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A ball is thrown vertically upward with a speed v from a height h metre above the ground. The time taken for the ball to hit ground iscorrect answer.: v/g[1 √1 √2hg/v×v]?how to solve? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is thrown vertically upward with a speed v from a height h metre above the ground. The time taken for the ball to hit ground iscorrect answer.: v/g[1 √1 √2hg/v×v]?how to solve?.
A ball is thrown vertically upward with a speed v from a height h metre above the ground. The time taken for the ball to hit ground iscorrect answer.: v/g[1 √1 √2hg/v×v]?how to solve? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A ball is thrown vertically upward with a speed v from a height h metre above the ground. The time taken for the ball to hit ground iscorrect answer.: v/g[1 √1 √2hg/v×v]?how to solve? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is thrown vertically upward with a speed v from a height h metre above the ground. The time taken for the ball to hit ground iscorrect answer.: v/g[1 √1 √2hg/v×v]?how to solve?.
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