
A body falling from the rest has a velocity v after it falls through a height h. The distance it has to fall down for its velocity to become double will be,
A. 8h
B. 6h
C. 3h
D. 5h
Answer
557.4k+ views
Hint: From the question you will get necessary quantities ready to be substituted. You could consider two cases, one the case height h at which the velocity becomes v and then apply Newton’s equation of motion. You could repeat the same for height h’ at which the velocity is doubled (2v) and then get the answer by comparing the two.
Formula used:
Newton’s equation of motion,
${{v}^{2}}-{{u}^{2}}=2as$
Complete answer:
In the question, we are given a body that falls from rest. This body gains velocity v after falling through a height h. We are asked to find the distance that the body has to fall down such that its velocity becomes double.
Let us note down all the quantities that are given in the question. It is said that the body is falling from rest, so it means the initial velocity of the body is zero.
$u=0m/s$
The final velocity on reaching a height h is $vm/s$ and also, the acceleration of a body under free fall is that due to gravity,
$a=g$
Now let us recall Newton’s equation of motion given by,
${{v}^{2}}-{{u}^{2}}=2as$
Substituting the given values,
${{v}^{2}}-0=2gh$
$\therefore h=\dfrac{{{v}^{2}}}{2g}$ ……………………………… (1)
Let h’ be the height the body reaches at which its velocity is doubled, then,
${{\left( 2v \right)}^{2}}-0=2gh'$
$\Rightarrow h'=\dfrac{4{{v}^{2}}}{2g}$
From (1),
$h'=4h$
The distance it has to travel from height h so as to double the velocity is given by,
$4h-h=3h$
It has to travel 3h distance more to gain double the velocity.
Therefore, at four times the given height, the velocity gets doubled.
Note:
Any body that has gravity as the only force acting upon it, then, the body is said to be in freefall. Clearly the body whose motion is discussed in the given question is under free fall. In vacuum, the body under free fall will accelerate with $g = 9.8m/s^2$ independent of the mass, but in the presence of air resistance, it is very different.
Formula used:
Newton’s equation of motion,
${{v}^{2}}-{{u}^{2}}=2as$
Complete answer:
In the question, we are given a body that falls from rest. This body gains velocity v after falling through a height h. We are asked to find the distance that the body has to fall down such that its velocity becomes double.
Let us note down all the quantities that are given in the question. It is said that the body is falling from rest, so it means the initial velocity of the body is zero.
$u=0m/s$
The final velocity on reaching a height h is $vm/s$ and also, the acceleration of a body under free fall is that due to gravity,
$a=g$
Now let us recall Newton’s equation of motion given by,
${{v}^{2}}-{{u}^{2}}=2as$
Substituting the given values,
${{v}^{2}}-0=2gh$
$\therefore h=\dfrac{{{v}^{2}}}{2g}$ ……………………………… (1)
Let h’ be the height the body reaches at which its velocity is doubled, then,
${{\left( 2v \right)}^{2}}-0=2gh'$
$\Rightarrow h'=\dfrac{4{{v}^{2}}}{2g}$
From (1),
$h'=4h$
The distance it has to travel from height h so as to double the velocity is given by,
$4h-h=3h$
It has to travel 3h distance more to gain double the velocity.
Therefore, at four times the given height, the velocity gets doubled.
Note:
Any body that has gravity as the only force acting upon it, then, the body is said to be in freefall. Clearly the body whose motion is discussed in the given question is under free fall. In vacuum, the body under free fall will accelerate with $g = 9.8m/s^2$ independent of the mass, but in the presence of air resistance, it is very different.
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