Kinematics Question 223
Question: A particle is thrown vertically upwards. If it’s velocity at half of the maximum height it’s 10 m/s, then maximum height attained by is (Take $ g=10 m/s^2 $) [CBSE PMT 2001, 2004]
Options:
A) 8 m
B) 10 m
C) 12 m
D) 16 m
Show Answer
Answer:
Correct Answer: B
Solution:
Let particle thrown with velocity u and it’s maximum height it’s H then $ H=\frac{u^{2}}{2g} $ When particle is at a height $ H/2 $ , then it’s speed is 10m/s From equation $ v^{2}=u^{2}-2gh $
$ {{(10)}^{2}}=u^{2}-2g( \frac{H}{2} )=u^{2}-2g\frac{u^{2}}{4g} $
$ \Rightarrow u^{2}=200 $ Maximum height
$ \Rightarrow H=\frac{u^{2}}{2g}=\frac{200}{2\times 10}=10\ m $
BETA
