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/s2) [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 $



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