Question
Question: A body is released from the top of a tower of height \(100m\). After \(2\) seconds it is stopped and...
A body is released from the top of a tower of height 100m. After 2 seconds it is stopped and then instantaneously released. What will be its height after the next 3 seconds? [take g=10m/s2]
(A) 40m
(B) 35m
(C) 45m
(D) 30m
Solution
Hint
To solve this question, we need to use the second kinematic equation of motion. As the body is dropped two times, we need to apply the equation also two times, to calculate the total downward displacement of the body in the two intervals. Then we need to subtract this displacement from the original height.
The formula used to solve this question is
⇒s=ut+21at2
⇒u=initial velocity, s= displacement, a= acceleration, and t=time
Complete step by step answer
As the body is released from the top, its initial velocity, u=0
Now, the vertical displacement covered by the body in 2 seconds is calculated using
⇒s=ut+21at2 (1)
Putting s=h1,u=0, a=g=10m/s2, and t=2s
⇒h1=0(2)+21(10)22
On solving, we get
⇒h1=20m
Now, as the body is stopped and then released, for the second displacement h2 of the body the initial velocity u=0
Now substituting s=h2 u=0, a=g=10m/s2, and t=3s in (1), we get
⇒h2=0(3)+21(10)32
On solving, we get
⇒h2=45m
Now, the total displacement covered by the body
⇒d=h1+h2
⇒d=20+45
On adding, we get
⇒d=65m
Now, the original height of the body h=100m
Therefore, the height of the body h′ = Original height − Total displacement
⇒h′=100−65
⇒h′=35m
So, the height of the body after the next 3 seconds is equal to 35m.
Hence, the correct answer is option (B), 35m
Note
The total vertical displacement 65m is not the final height of the body after the next 3 seconds. It is the downward displacement of the body. Do not forget to subtract this displacement from the original height. It is a common mistake that we forget to make this subtraction in the end and conclude the downward displacement of the body as its final height.