A particle is thrown vertically upwards. If it velocity at half of the maximum height is $10\ m/sec$, then maximum height attained by it is (Take $g = 10\ m/\sec^{2}\ $)

Question

Dr. Abubacker · Accepted Answer

(b) Let particle thrown with velocity $u$ and its maximum height is $H$ then $H = \frac{u^{2}}{2g}$
When particle is at height $H/2$, then its speed is $10\ m/s$
From equation $v^{2} = u^{2} - 2gh$
$(10)^{2} = u^{2} - 2g\left( \frac{H}{2} ight) = u^{2} - 2g\frac{u^{2}}{4g}\  \Rightarrow u^{2} = 200$
Maximum height $\Rightarrow H = \frac{u^{2}}{2g} = \frac{200}{2 	imes 10} = 10\ m$

Q. A particle is thrown vertically upwards. If it velocity at half of the maximum height is $10\ m/sec$, then maximum height attained by it is (Take $g = 10\ m/\sec^{2}\ $)

Explanation

Reviewed by Dr. Abubacker

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