A body is thrown horizontally with an initial velocity of `10m//s` from a cliff. At what time (in sec.) will the velocity vector make an angle of `45^(@)` with horizontal?

A projectile is thrown with a velocity of 20 m//s, at an angle of 60^(@) with the horizontal. After how much time the velocity vector will make an angle of 45^(@) with the horizontal (in upward direction) is (take g= 10m//s^(2))-

A projectile is thrown with a velocity of 20 m//s, at an angle of 60^(@) with the horizontal. After how much time the velocity vector will make an angle of 45^(@) with the horizontal (in upward direction) is (take g= 10m//s^(2))-

A particle is projected from horizontal making an angle 60^@ with initial velocity 50m/s . The time taken by the particle to make an angle of 45^@ from the horizontal is

From the the top of a tower 19.6 m high, a ball is thrown horizontally. If the line joining the point of projection to the point where it hits the ground makes an angle of 45^(@) with the horizontal, then the initial velocity of the ball is:-

From the the top of a tower 19.6 m high, a ball is thrown horizontally. If the line joining the point of projection to the point where it hits the ground makes an angle of 45^(@) with the horizontal, then the initial velocity of the ball is:-

From the the top of a tower 19.6 m high, a ball is thrown horizontally. If the line joining the point of projection to the point where it hits the ground makes an angle of 45^(@) with the horizontal, then the initial velocity of the ball is:-

From the top of a tower 20m high, a ball is thrown horizontally. If the line joining the point of projection to the point where it hits the ground makes an angle of 45^(@) with the horizontal, then the initial velocity of the ball is:

A projectile is thrown in the upward direction making an angle of 60^@ with the horizontal with a velocity of 140 ms^-1 . Then the time after which its velocity makes an angle 45^@ with the horizontal is (Acceleration due to gravity g=10 ms^-2 )