A particle aimed at a target projected with an angle `15^(@)` with the horizontal is short of the target by 10m .If projected with an angle of `45^(@)` is away from the target by 15m then the angle of projection to hit the target is

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

A bullet is fired at an angle of 15^(@) with the horizontal and it hits the ground 3 km away. Can we hit a target at a distance of 7 km by just adjusting its angle of projection ?

A bullet fired at an angle of 30^(@) with the horizontal hits the ground 3 sqrt 3 km away. Can wr hit a target at a distance of 6 sqrt 2km by adjustion its angle of projection?

A ball projected with a velocity of 28m/sec has a horizontal range 40m. Find the two angles of projection.

A bullet is fired at an angle of 15^@ with the horizontal and hits the ground 6 km away. Is it possible to hit a target 10 km away by adjusting the angle of projection assuming the initial speed to be the same ?

A projectile is aimed at a mark on the horizontal plane through the point of projection and falls 12 m short when the angle of projection is 15^(@) , white it overshoots the mark by 24 m when the angle of projection is 45^(@) . Find the angle of projection to hit the mark.

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 of 10m/s at an angle of 15° with horizontal.The horizontal range will be (g = 10m/s^2)

A projectile is fired at an angle of 45^(@) with the horizontal. Elevation angle of the projection at its highest point as seen from the point of projection is