A body is projected at an angle of `30^(@)` with the horizontal and with a speed of `30ms^(-1)`. What is the angle with the horizontal after 1.5 second? `(g=10ms^(-2))`

A body is projected at an angle of 30^@ with the horizontal and with a speed of 30 ms^-1 . What is the angle with the horizontal after 1.5 s ? (g = 10 ms^-2) .

A particle is projected at an angle theta =30^@ with the horizontal, with a velocity of 10ms^(-1) . Then

A body is projected at angle 30^(@) to horizontal with a velocity 50 ms^(-1) .Its time of flight is

A ball is projected at an angle 60^(@) with the horizontal with speed 30 m/s. What will be the speed of the ball when it makes an angle 45^(@) with the horizontal ?

A particle is projected with speed 'u' at an angle 60^@ with horizontal, then find the speed when its angle with horizontal changes to 30^@ .

A stone is projected with speed of 50 ms^(-1) at an angle of 60^(@) with the horizontal. The speed of the stone at highest point of trajectory is

A body is projected at an angle 60^@ with horizontal with kinetic energy K. When the velocity makes an angle 30^@ with the horizontal, the kinetic energy of the body will be

A projectile is projected in the upward direction making an angle of 45^@ with horizontal direction with a velocity of 150 ms^(-1) . Then the time after which its inclination with the horizontal is 30^@ is (g=10 ms^(-2)).