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A stone is released from the top of the tower. it covers 24.5 m distance in the last second of its journey. The time, for which stone is air is
A 3 sec
B 9 sec
C 1 sec
D 5 sec

Last updated date: 05th Mar 2024
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IVSAT 2024
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Hint for these types of questions, we use the Newton’s equation of motion i.e. $s = ut + \dfrac{1}{2}a{t^2}$, here firstly u = 0 then we can find the value of s. after that it is given that in the last second of the journey i.e. t = t-1, so replace the value of t by t-1 in the value of s and then equate it with 24.5 m, we will get the required result.

Step by step solution
As, here it is given that the stone is released from the top of the tower, it means initial velocity is zero, and let the stone travel the distance s from the top.
Using Newton’s equation of motion is
$s = ut + \dfrac{1}{2}a{t^2}$
As, u = 0 and as stone is falling under the gravity then take a = g, we get
$ \Rightarrow s = \dfrac{1}{2}g{t^2}$…………. (1)
As, $g = 9.8m{s^{ - 2}}$ then
$ \Rightarrow s = 4.9{t^2}$ …………………. (2)
Now, as it is given that the distance travel by the stone in last second i.e. when t = t-1 seconds is 24.5m, therefore we can substitute the t = t-1 in equation (2) , we get
$ \Rightarrow s = 4.9{\left( {t - 1} \right)^2}$ …………………. (3)
Subtracting equation (3) from equation (2), and equate it to 24.5 m, we get
$ \Rightarrow 4.9{t^2} - 4.9{\left( {t - 1} \right)^2} = 24.5$
On solving the above equation, we get
$ \Rightarrow {t^2} - {\left( {t - 1} \right)^2} = 5$
$ \Rightarrow 2t - 1 = 5$
$ \Rightarrow t = 3\sec $
Hence, the time for which stone will remain in air is 3 second

Therefore, option A is correct.

Note: In this equation it must be noticed that when the object is released from the top of the tower it means its initial velocity is equal to zero. After substituting the value we will get the equation.
We have three Newton’s equation of motion i.e. $v = u + at$, ${v^2} - {u^2} = 2as$, $s = ut + \dfrac{1}{2}a{t^2}$
Which equation we have to use this will depend upon the conditions given in the question.