Kinematics Question 408
Question: A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18s. What is the value of v? $ (takeg=10m/s^{2}) $
Options:
A) 75 m/s
B) 55 m/s
C) 40 m/s
D) 60 m/s
Show Answer
Answer:
Correct Answer: A
Solution:
Clearly distance moved by 1st ball in 18s = distance moved by 2nd ball in 12s.
Now, distance moved in 18 s by 1st ball $ =\frac{1}{2}\times 10\times 18^{2}=90\times 18=1620\text{ m } $
distance moved in 12 s by 2nd ball $ =ut+\frac{1}{2}gt^{2}\text{ }\therefore \text{1620=12v+5}\times 144 $
$ \Rightarrow \text{v=135-60=75 m}{{s}^{-1}} $
BETA
