At a height `0.4 m` from the ground the velocity of a projectile in vector form is `vec v = (6 hat i+ 2 hat j) ms ^-1`. The angle of projection isA. `45^@`B. `60^@`C. `30^@`D. `tan^-1 (3//4)`

At a height `0.4 m` from the ground the velocity of a projectile in ve

1.	At a height `0.4 m` from the ground the velocity of a projectile in vector form is `vec v = (6 hat i+ 2 hat j) ms ^-1`. The angle of projection isA. `45^@`B. `60^@`C. `30^@`D. `tan^-1 (3//4)`
Answer» Correct Answer - C ( c) `v^2 = u^2 - 2 g h` or `u^2 = v^2 + 2gh` or `u_x^2 + u_y^2 = v_x^2 + v_y^2 + 2gh , u_x = v_x` So, `u_y^2 = v_y^2 + 2gh` or `u_y^2 = (2)^2 + 2 xx 10 xx 0.4 = 12` `u_y = (sqrt(12)) = 2 (sqrt(3))^-1, u_x = v_x = 6 ms^-1` `tan theta = (u_y)/(u_x) = (2sqrt(3))/(6) = (1)/(sqrt(3)) rArr theta = 30^@`.

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At a height `0.4 m` from the ground the velocity of a projectile in vector form is `vec v = (6 hat i+ 2 hat j) ms ^-1`. The angle of projection isA. `45^@`B. `60^@`C. `30^@`D. `tan^-1 (3//4)`

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