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A stone with weight w is thrown vertically upward into air from ground level with initial speed v0. If a constant force F due to air drag acts on the stone throughout its flight. The maximum height attained by the stone is 1) h= vo²/ 2g(1 f/w) 2) h= vo²/2g(1-f/w) 2) h=vo²/2g(1 w/f) 3) h= vo²/2g(1-w/f)?
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A stone with weight w is thrown vertically upward into air from ground...
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A stone with weight w is thrown vertically upward into air from ground...
Understanding the Problem
We are given a stone with weight w that is thrown vertically upward into the air from ground level with an initial speed v0. We also know that a constant force F due to air drag acts on the stone throughout its flight. We need to determine the maximum height attained by the stone.
Solving the Problem
To find the maximum height attained by the stone, we need to analyze the forces acting on it and apply the equations of motion.
Forces Acting on the Stone
- The weight of the stone acts downwards with a magnitude of w.
- The force due to air drag acts in the opposite direction of motion.
Initial Velocity and Acceleration
- The initial velocity of the stone is v0.
- The acceleration due to gravity is denoted by g and acts downwards.
Using Equations of Motion
We can use the equations of motion to analyze the vertical motion of the stone.
We know that the final velocity of the stone at its maximum height is 0, as it comes to a stop before falling back down. Using the equation of motion:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement, we can find the displacement of the stone.
At the maximum height, the final velocity of the stone is 0, so the equation becomes:
0 = v0^2 + 2as
Simplifying the equation, we get:
s = -v0^2 / (2a)
Since the stone is moving upwards, the acceleration due to gravity acts in the opposite direction. Therefore, we substitute -g for a:
s = -v0^2 / (2(-g))
Simplifying further:
s = v0^2 / (2g)
Considering Air Drag
Now, we need to consider the effect of the constant force F due to air drag on the stone's motion.
The force due to air drag acts in the opposite direction of motion and depends on the velocity of the stone. Let's assume that the force due to air drag is proportional to the velocity of the stone, which means:
F = kf
where k is a constant of proportionality and f is the velocity of the stone.
The work done by the force due to air drag is given by:
W = F * s
Substituting the value of F and s, we get:
W = kf * (v0^2 / (2g))
Since the work done is equal to the change in kinetic energy, we have:
W = w * g * h
where w is the weight of the stone and h is the maximum height attained.
Calculating the Maximum Height
Equating the expressions for work done, we get:
kf * (v0^2 / (2g)) = w * g * h
Simplifying the equation, we find:
h = v0^2 / (2g(1 - f/w))
Therefore, the correct option is:
2) h = v0^2 / (2g(1 - f/w))
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A stone with weight w is thrown vertically upward into air from ground level with initial speed v0. If a constant force F due to air drag acts on the stone throughout its flight. The maximum height attained by the stone is 1) h= vo²/ 2g(1 f/w) 2) h= vo²/2g(1-f/w) 2) h=vo²/2g(1 w/f) 3) h= vo²/2g(1-w/f)?
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A stone with weight w is thrown vertically upward into air from ground level with initial speed v0. If a constant force F due to air drag acts on the stone throughout its flight. The maximum height attained by the stone is 1) h= vo²/ 2g(1 f/w) 2) h= vo²/2g(1-f/w) 2) h=vo²/2g(1 w/f) 3) h= vo²/2g(1-w/f)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A stone with weight w is thrown vertically upward into air from ground level with initial speed v0. If a constant force F due to air drag acts on the stone throughout its flight. The maximum height attained by the stone is 1) h= vo²/ 2g(1 f/w) 2) h= vo²/2g(1-f/w) 2) h=vo²/2g(1 w/f) 3) h= vo²/2g(1-w/f)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone with weight w is thrown vertically upward into air from ground level with initial speed v0. If a constant force F due to air drag acts on the stone throughout its flight. The maximum height attained by the stone is 1) h= vo²/ 2g(1 f/w) 2) h= vo²/2g(1-f/w) 2) h=vo²/2g(1 w/f) 3) h= vo²/2g(1-w/f)?.
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