1.

A soft drink bottle is dropped by a water from the top of a hotel whose height is 100 m. It takes T second to reach the ground. Where will it be at T/2 second?A. 50 m from topB. 75 m from bottomC. 25 m from bottomD. 40 m from top

Answer»

Height of Tower = 100 m

Time taken to reach ground = T seconds

Acceleration = a = g = 9.8 ms-2

Initial Speed = u = 0

Final Speed = v

Now, we know that,

S = ut + \(\frac{1}{2}\)at2

100 = \(\frac{1}{2}\)gT2 At t = T and S = 100 m..........1

Let t = \(\frac{T}{2}\) and s = x

Then, x = \(\frac{1}{2}\)a\(\frac{T^2}{2}\)..........2

So, dividing equation 1 by 2, we get,

\(\frac{100}{\times}\) = 4

x = 25 m

So, 25 m from the top at t = \(\frac{T}{2}\) seconds or (100 - 25)m, that is 75 m from the bottom at \(\frac{T}{2}\)seconds.

Hence, option B is correct.


Discussion

No Comment Found

Related InterviewSolutions