A particle is projected from horizontal making an angle of `53^(@)` with initial velocity `100 ms^(-1)`. The time taken by the particle to make angle `45^(@)` from horizontal is

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

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

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

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 )

A particle is projected at an angle of 45^@ with a velocity 9.8ms^-1 . What is horizontal range.

A particle is projected with velocity 50 m/s at an angle 60^(@) with the horizontal from the ground. The time after which its velocity will make an angle 45^(@) with the horizontal is

A particle is projected with velocity 50 m/s at an angle 60^(@) with the horizontal from the ground. The time after which its velocity will make an angle 45^(@) with the horizontal is

A particle is projected from the ground at an angle of 60^(@) with horizontal at speed u = 20 m//s. The radius of curvature of the path of the particle, when its velocity. makes an angle of 30^(@) with horizontal is : (g=10 m//s^(2)

A projectile is projected from horizontal with velocity (u) making an angle 45^(@) with the horizontal direction. Find the distance of the highest point of the projectile from its starting point.