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Question: A particle is thrown vertically upwards. If its velocity at half of the maximum height is 10 m/s, th...

A particle is thrown vertically upwards. If its velocity at half of the maximum height is 10 m/s, then maximum height attained by it is (Take g=10g = 10 m/s2)

A

8 m

B

10 m

C

12 m

D

16 m

Answer

10 m

Explanation

Solution

Let particle thrown with velocity u and its maximum height is H then H=u22gH = \frac{u^{2}}{2g}

When particle is at a heightH/2H/2, then its speed is 10 m/s

From equation v2=u22ghv^{2} = u^{2} - 2gh

(10)2=u22g(H2)=u22gu24gu2=200(10)^{2} = u^{2} - 2g\left( \frac{H}{2} \right) = u^{2} - 2g\frac{u^{2}}{4g} \Rightarrow u^{2} = 200

Maximum height H=u22g=2002×10=106mum\Rightarrow H = \frac{u^{2}}{2g} = \frac{200}{2 \times 10} = 10\mspace{6mu} m