A particle is projected from the ground with velocity `28.28ms^(-1)`at an angle of `45^(@)`to horizontal.The particle will be at a height of "15m above the ground twice during its motion.What is the time gap,in seconds,between those two instants?

A particle is projected from ground with velocity 40sqrt2m at 45^(@) .At time t=2s

A particle is projected from ground with velocity 40sqrt(2)m//s at 45^(@) . At time t=2s

A particle is projected form ground with velocity u ar angle theta from horizontal. Match the following two columns.

A particle is projected with a velocity of 20 ms^(-1) at an angle of 60^(@) to the horizontal. The particle hits the horizontal plane again during its journey. What will be the time of impact ?

A particle is projected from ground with some initial velocity making an angle of 45^(@) with the horizontal. It reaches a height of 7.5 m above the ground while it travels a horizontal distance of 10 m from the point of projection. The initial speed of the projection is

A particle is projected from ground with velocity 20(sqrt2) m//s at 45^@ . At what time particle is at height 15 m from ground? (g = 10 m//s^2)

A particle is projected from the ground with a kinetic energy E at an angle of 60° with the horizontal. Its kinetic energy at the highest point of its motion will be

A ball is projected from ground with a speed of 20ms^(-1) at an angle of 45^(@) with horizontal.There is a wall of 25m height at a distance of 10m from the projection point. The ball will hit the wall at a height of

A particle is projected with a velocity 10 m//s at an angle 37^(@) to the horizontal. Find the location at which the particle is at a height 1 m from point of projection.