A particle falling from rest under gravity covers a height H in 5 seconds, if it continues falling, then the next distance H will be covered in what time?
(A) 2s
(B) 3s
(C) 4s
(D) 5s
Answer
261.6k+ views
Hint The given terms are H height and time period t of 5 seconds. It is also said that the particle is experiencing free fall from rest under gravity g. Now, using the given terms, apply a second equation of motion to identify the time period when the particle covers the next H distance.
Complete Step By Step Solution
It is given that an article is kept at a certain height and experiences free fall under gravity g. Now, it is given that the particle covers a distance H at 5 seconds of falling. Applying the second law of motion, we get,
\[s = ut + \dfrac{1}{2}g{t^2}\], where s is the displacement of the body from rest, u is the initial velocity of the object, t is the time period of the object under free fall.
Now in our first case it is given that the particle covers H distance in 5 seconds. This is given by
\[H = (0) \times 5 + \dfrac{1}{2}g{(5)^2}\]
\[ \Rightarrow H = \dfrac{{25g}}{2}\]
Now, the particle again drops another height H, in an unknown time period t. This is represented by the formula
\[2H = u \times t + \dfrac{1}{2}g{t^2}\], since displacement of the second drop is twice the height H from the rest position.
From the previous equation, we found out H value. Substituting this in the above equation we get,
\[ \Rightarrow 2(\dfrac{{25g}}{2}) = 0 \times t + \dfrac{1}{2}g{t^2}\]
Cancelling out common terms we get,
\[ \Rightarrow 50 = {t^2}\]
\[ \Rightarrow t \simeq 7\sec \]
Now we know for a total height of 2H, the particle takes 7 seconds to cross. Now, for the second H , the time it takes from to drop from H to 2H is ,
\[ \Rightarrow {t_{2H}} = {t_{0 - H}} + {t_{H - 2H}}\]
\[ \Rightarrow {t_{2H}} - {t_{0 - H}} = {t_{H - 2H}}\]
\[ \Rightarrow 7 - 5 = {t_{H - 2H}}\]
\[ \Rightarrow {t_{H - 2H}} = 2s\]
Hence, Option (a) is the right answer for the given question.
Note Equations of motion are generic equations that are used to describe the nature and behavior of any system with regards to motion constraints such as velocity, speed etc., as a function of time. These are dynamic variables and are vector quantities that change with direction and magnitude.
Complete Step By Step Solution
It is given that an article is kept at a certain height and experiences free fall under gravity g. Now, it is given that the particle covers a distance H at 5 seconds of falling. Applying the second law of motion, we get,
\[s = ut + \dfrac{1}{2}g{t^2}\], where s is the displacement of the body from rest, u is the initial velocity of the object, t is the time period of the object under free fall.
Now in our first case it is given that the particle covers H distance in 5 seconds. This is given by
\[H = (0) \times 5 + \dfrac{1}{2}g{(5)^2}\]
\[ \Rightarrow H = \dfrac{{25g}}{2}\]
Now, the particle again drops another height H, in an unknown time period t. This is represented by the formula
\[2H = u \times t + \dfrac{1}{2}g{t^2}\], since displacement of the second drop is twice the height H from the rest position.
From the previous equation, we found out H value. Substituting this in the above equation we get,
\[ \Rightarrow 2(\dfrac{{25g}}{2}) = 0 \times t + \dfrac{1}{2}g{t^2}\]
Cancelling out common terms we get,
\[ \Rightarrow 50 = {t^2}\]
\[ \Rightarrow t \simeq 7\sec \]
Now we know for a total height of 2H, the particle takes 7 seconds to cross. Now, for the second H , the time it takes from to drop from H to 2H is ,
\[ \Rightarrow {t_{2H}} = {t_{0 - H}} + {t_{H - 2H}}\]
\[ \Rightarrow {t_{2H}} - {t_{0 - H}} = {t_{H - 2H}}\]
\[ \Rightarrow 7 - 5 = {t_{H - 2H}}\]
\[ \Rightarrow {t_{H - 2H}} = 2s\]
Hence, Option (a) is the right answer for the given question.
Note Equations of motion are generic equations that are used to describe the nature and behavior of any system with regards to motion constraints such as velocity, speed etc., as a function of time. These are dynamic variables and are vector quantities that change with direction and magnitude.
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