If a projectile is fired at an angle `theta` with the vertical with velocity u, then maximum height attained is given by:-
A. `(u^(2)cos theta)/(2g)`
B. `(u^(2)sin^(2)theta)/(2g)`
C. `(u^(2)sin^(@)theta)/(g)`
D. `(u^(2)cos^(2)theta)/(2g)`

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Correct Answer - D
`H=(u_(y)^(2))/(2g)=(u^(2)cos^(2)theta)/(2g)`

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