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A particle is thrown vertically upward. ...

A particle is thrown vertically upward. Its velocity at half of the height is 10 m/s. Then the maximum height attained by it : -
`(g=10 m//s^2)`

Text Solution

Verified by Experts

Let initial velocity of particle = u.
Half of the maximum height = `(1)/(2) (u^(2))/(2g) =(u^(2))/(4g)`
At half of maximum height, `v=10 ms^(-1)`
Using, `v^(2)=u^(2)+2as" or "100-u^(2) =2 (-g) (u^(2))/(4g) =-(u^(2))/(2)" or "(u^(2))/(2) =100`
? Maximum height attained `=(u^(2))/(2g) =(u^(2))/(2) xx (1)/(g) =(100)/(10) =10 m`
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