Question
Question: A body is released from the top of a tower of height H m. after 2s it is stopped and then instantane...
A body is released from the top of a tower of height H m. after 2s it is stopped and then instantaneously released. What will be its height after the next 2s?
A) (H-5)m
B) (H-10)m
C) (H-20)m
D) (H-40)m
Solution
Apply the second kinematic equation of motion to solve this problem. As mentioned in the question the body dropped two times, so we need to apply the equation also two times, to calculate the downward displacement of the body in the two intervals. In the next step, we need to subtract this displacement from the original height.
Formula used: s=ut+21at2
Complete step by step answer:
s=ut+21at2------(1)
u is the initial velocity
s is the displacement
a is the acceleration
t is the time
when body is released from the top its initial velocity is zero means u=0 now, the vertical displacement covered by the body in 2s s=$h_{1}$, u=0, a=g=10m/$s^{2}$, t=2s by putting these value we get ${{h}_{1}}=0(2)+\dfrac{1}{2}(10){{2}^{2}}$ $\Rightarrow {{h}_{1}}=10\times 2=20m$ Now, body is stopped and then released for the second displacement $h_{2}$ of the body. Now we take, initial velocity u=0 Now substituting s=$h_{2}$,u=0 a=g=10m/$s^{2}$, and t=2s these value in equation number (1) ${{h}_{2}}=0(2)+\dfrac{1}{2}(10){{2}^{2}}$ $\Rightarrow {{h}_{2}}=10\times 2=20m$ Now, the total displacement covered by the body d=$h_{1}$+$h_{2}$ by substituting the value we get d=20+20 by adding we get d=40m the original height of the body h=Hm $\therefore $ the height of the body h
=Original height – Total displacement
⇒h`=Hm-40m
So, the height of the body after the next 2s is equal to (H-40)m
So, the correct answer is “Option D”.
Note: In this problem is the downward displacement of the body so don't forget to subtract this displacement from the original height.