Kinematics Question 413
Question: A ball is dropped from the top of a tower of height 100 m and at the same time another ball is projected vertically upwards from ground with a velocity $ 25m{{s}^{-1}} $ . Then the distance from the top of the tower, at which the two balls meet it’s
Options:
A) 68.4 m
B) 48.4 m
C) 18.4 m
D) 78.4 m
Show Answer
Answer:
Correct Answer: D
Solution:
Let the two balls P and Q meet at height x m from the ground after time t s from the start. We have to find distance, $ BC=( 100-x ) $
For ball P $ \text{S=x m, u=25 m}{{s}^{-1}},a=-g $ From $ S=ut+\frac{1}{2}at^{2} $
$ x=25t-\frac{1}{2}at^{2} $ ………. (i)
For ball Q $ \text{S= }( 100-x )\text{ m, u = 0, a =g} $
$ \therefore 100-x=0+\frac{1}{2}gt^{2} $ ………. (ii)
Adding eqn. (i) and (ii), we get $ \text{100=25t or t=4s} $
From eqn. (i), $ \text{x=25}\times 4-\frac{1}{2}\times 9.8\times {{( 4 )}^{2}}=21.6\text{ m } $
Hence distance from the top of the tower $ \text{= (100-x)m=(100-21}\text{.6m)=78}\text{.4m} $
BETA
